Chapter 6 - Applying Newton's Laws - Modeling with Physics

6.1Overview¶

In this chapter, we take a closer look at how to use Newton’s Laws to build models to describe motion. Whereas the previous chapter was focused on identifying the forces that are acting on an object, this chapter focuses on using those forces to describe the motion of the object.

Newton’s Laws are meant to describe “point particles”, that is, objects that can be thought of as a point and thus have no orientation. A block sliding down a hill, a person on a merry-go-round, a bird flying through the air can all be modelled as point particles, as long as we do not need to model their orientation. In all of these cases, we can model the forces on the object using a free-body diagram as the location of where the forces are applied on the object do not matter. In later chapters, we will introduce the tools required to apply Newton’s Second Law to objects that can rotate, where we will see that the location of where a force is exerted matters.

6.2Statics¶

When using Newton’s Laws to model an object, one can identify two broad categories of situations: static and dynamic. In static situations, the acceleration of the object is zero. By Newton’s Second Law, this means that the vector sum of the forces (and torques, as we will see in a later chapter) exerted on an object must be zero. In dynamic situations, the acceleration of the object is non-zero.

For static problems, since the acceleration vector is zero, we can choose a coordinate system in a way that results in as many forces as possible being aligned with the axes (so that we minimize the number of forces that we need to break up into components).

Example 6.1

You push horizontally with a force $\vec F$ on a box of mass $m$ that is resting against a vertical wall, as shown in Figure 1. The coefficient of static friction between the wall and the box is $\mu_s$ . What is the minimum magnitude of the force that you must exert for the box to remain stationary?

Figure 1:A horizontal force exerted on box that is resting against a wall.

Solution

Since the acceleration of the box is zero, the vector sum of the forces exerted on the box is zero. We start by identifying the forces exerted on the box; these are:

$\vec F$ , the horizontal force that you exert on the box.
$\vec F_g$ , the weight of the box, with magnitude $mg$ .
$\vec N$ , a normal force exerted by the wall on the box. The force is in the horizontal direction, in the opposite direction to $\vec F$ .
$\vec f_s$ , a vertical force of static friction between the wall and the box. The force points upwards as the “impeding motion” of the block is downwards. The force will have at most a magnitude of $f_s\leq\mu_s N$ , since the force of static friction depends on the other forces exerted on the object.

The forces are shown in the free-body diagram in Figure 2, along with our choice of coordinate system which was chosen so that all forces are either in the $x$ or $y$ direction.

Figure 2:Free-body diagram of the forces exerted on the box.

The $x$ component of Newton’s Second Law is:

\begin{align*} \sum F_x = F - N &=0\\ \therefore N = F \end{align*}

(1)

which tells us that the normal force exerted by the wall has the same magnitude as the applied force, $\vec F$ . The $y$ component of Newton’s Second Law is:

\begin{align*} \sum F_y = f_s - F_g &=0\\ \therefore f_s -mg &=0\\ \therefore f_s = mg\\ \end{align*}

(2)

which tells us that the force of friction must have the same magnitude as the weight. This makes sense, since they are the only forces with components in the $y$ direction, and thus, they must cancel each other out.

The force of friction will be less than or equal to $\mu_sN$ , and thus less than or equal to $\mu_s F$ , since $\vec F$ and $\vec N$ have the same magnitude (from the $x$ component of Newton’s Second Law). Furthermore, since $f_s=mg$ , we can write:

\begin{align*} f_s &\leq \mu_s F\\ \therefore mg &\leq \mu_s F\\ \therefore \frac{mg}{\mu_s} &\leq F \end{align*}

(3)

which gives us the condition that $F\geq mg/\mu_s$ , and thus the minimum magnitude of $F$ in order to keep the box from sliding down.

Although we used the lesser than or equal to sign in the above equations, we could have used an equal sign if we were confident that the force of friction has its maximal magnitude, $f_s=\mu_sN$ . The maximal magnitude of the force of friction is proportional to the force that we exert (since $N=F$ ); if we want to exert the least amount of force $F$ , then we need the force of friction to be equal to its maximal magnitude which needs to be equal to the weight of the box.

Discussion: This model for the minimal required force makes sense because:

The dimension of $mg/\mu_s$ is force.
If the mass of the box is increased, then one needs to push harder against the box to keep it up.
If the coefficient of static friction, $\mu_s$ , is increased, one does not need to push as hard.

6.3Linear motion¶

We can describe the motion of an object whose velocity vector does not continuously change direction as “linear” motion. For example, an object that moves along a straight line in a particular direction, then abruptly changes direction and continues to move in a straight line can be modelled as undergoing linear motion over two different segments (which we would model individually). An object moving around a circle, with its velocity vector continuously changing direction, would not be considered to be undergoing linear motion. For example, paths of objects undergoing linear and non-linear motion are illustrated in Figure 3.

Figure 3:(Left:) Displacement vectors for an object undergoing three segments that can each be modelled as linear motion. (Right:) Path of an object whose velocity vector changes continuously and cannot be considered as linear motion.

When an object undergoes linear motion, we always model the motion of the object over straight segments separately. Over one such segment, the acceleration vector will be co-linear with the displacement vector of the object (parallel or anti-parallel - note that the acceleration can change direction as it would from a spring force, but will always be co-linear with the displacement).

Example 6.2

A block of mass $m$ is placed at rest on an incline that makes an angle θ with respect to the horizontal, as shown in Figure 4. The block is nudged slightly so that the force of static friction is overcome and the block starts to accelerate down the incline. At the bottom of the incline, the block slides on a horizontal surface.

The coefficient of kinetic friction between the block and the incline is $\mu_{k1}$ , and the coefficient of kinetic friction between the block and horizontal surface is $\mu_{k2}$ . If one assumes that the block started at rest a distance $L$ from the bottom of the incline, how far along the horizontal surface will the block slide before stopping?

Figure 4:A block slides down an incline before sliding on a flat surface and stopping.

6.3.1Modelling situations where forces change magnitude¶

So far, the models that we have considered involved forces that remained constant in magnitude. In many cases, the forces exerted on an object can change magnitude and direction. For example, the force exerted by a spring changes as the spring changes length or the force of drag changes as the object changes speed. In these case, even if the object undergoes linear motion, we need to break up the motion into many small segments over which we can assume that the forces are constant. If the forces change continuously, we will need to break up the motion into an infinite number of segments and use calculus.

Consider the block of mass $m$ that is shown in Figure 7, which is sliding along a frictionless horizontal surface and has a horizontal force $\vec F(x)$ exerted on it. The force has a different magnitude in the three segments of length $\Delta x$ that are shown. If the block starts at position $x=x_0$ axis with speed $v_0$ , we can find, for example, its speed at position $x_3=3\Delta x$ , after the block travelled through the three segments.

Figure 7:A block being pushed along a frictionless horizontal surface with a force that changes.

The horizontal force, $\vec F$ , exerted on the block can be written as:

\vec F (x)= \begin{cases} F_1\hat x & x<\Delta x \quad \text{(segment 1)}\\ F_2\hat x & \Delta x \leq x< 2\Delta x \quad \text{(segment 2)}\\ F_3\hat x & 2\Delta x \leq x\quad \text{(segment 3)} \end{cases}

(12)

as it depends on the location of the block. To find the speed of the block at the end of the third segment, we can model each segment separately. The forces exerted on the block are the same in each segment:

$\vec F_g$ , its weight, with magnitude $mg$ .
$\vec N$ , a normal force exerted by the ground.
$\vec F(x)$ , an applied force that changes magnitude with position and is different in the three different segments.

The forces are illustrated in the free-body diagram show in Figure 8.

Free-body diagram for the block shown in [](#fig:applyingnewtonslaws:blockvaryingforce). — Figure 8:Free-body diagram for the block shown in Figure 7.

Newton’s Second Law can be used to determine the acceleration of the block for each of the three segments, since the forces are constant within one segment. For all three segments, the $y$ component of Newton’s Second Law just tells us that the normal force exerted by the ground is equal in magnitude to the weight of the block. The $x$ component of Newton’s Second Law gives the acceleration:

\sum F_x = F_i = ma_i

(13)

where we have used the index $i$ to indicate which segment the block is in ( $i$ can be 1, 2 or 3). The acceleration of the block in segment $i$ is given by:

a_i = \frac{F_i}{m}

(14)

If the speed of the block is $v_0$ at the beginning of segment 1 ( $x=x_0$ ), we can find its speed at the end of segment 1 ( $x=x_1$ ), $v_1$ , using kinematics and the fact that the acceleration in segment 1 is $a_1$ :

\begin{align*} v_1^2-v_0^2 &= 2a_1(x_1 - x_0)\\ v_1^2 &=v_0^2+ 2a_1\Delta x\\ \therefore v_1^2 &=v_0^2+2\frac{F_1}{m}\Delta x \end{align*}

(15)

We can now easily find the speed at the end of segment 2 ( $x=x_2$ ), $v_2$ , since we know the speed at the beginning of segment 2 ( $x_1$ , $v_1$ ) and the acceleration $a_2$ :

\begin{align*} v_2^2 -v_1^2 &= 2a_2(x_2 - x_1)\\ \therefore v_2^2 &= v_1^2 + 2a_2\Delta x\\ &=v_0^2+ 2\frac{F_1}{m}\Delta x + 2\frac{F_2}{m}\Delta x \end{align*}

(16)

It is easy to show that the speed at the end of the third segment is:

v_3^2 = v_0^2+ 2\frac{F_1}{m}\Delta x + 2\frac{F_2}{m}\Delta x +2\frac{F_3}{m}\Delta x

(17)

If there were $N$ segments, with the force being different in each segment, we could use the summation notation to write:

\begin{align*} v_N^2 &= v_0^2 + 2\sum_{i=1}^{i=N} \frac{F_i}{m}\Delta x \end{align*}

(18)

Finally, if the magnitude of the force varied continuously as a function of $x$ , $\vec F(x)$ , we would model this by taking segments whose length, $\Delta x$ , tends to zero (and we would need an infinite number of such segments). For example, if we wanted to know the speed of the object at position $x=X$ along the $x$ axis, with a force that was given by $\vec F(x)=F(x)\hat x$ , if the object started at position $x_0$ with speed $v_0$ , we would take the following limit:

v^2 = v_0^2 + \lim_{\Delta x \to 0} 2\sum_{i=1}^{i=N} \frac{F(x)}{m}\Delta x

(19)

where $\Delta x = \frac{X}{N}$ so that as $\Delta x\to 0$ , $N\to\infty$ . Of course, integrals are the exact tool that allow us to evaluate the sum in this limit:

\lim_{\Delta x \to 0} 2\sum_{i=1}^{i=N} \frac{F_i}{m}\Delta x =2 \int_{x_0}^{X}\frac{F(x)}{m}dx

(20)

and the speed at position $x=X$ is given by:

v^2 = v_0^2 + 2 \int_{x_0}^{X}\frac{F(x)}{m}dx

(21)

Naturally, we can find the above result starting directly from calculus. If the component of the (net) force in the $x$ direction is given by $F(x)$ , then the acceleration is given by $a(x) = \frac{F(x)}{m}$ . The velocity is related to the acceleration:

\begin{align*} a(x) &= \frac{dv}{dt}\\ \therefore dv &= a(x)dt\\ \end{align*}

(22)

We cannot simply integrate the last equation to find that $v=\int a(x)dt$ because the acceleration is given as a function of position, $a(x)$ , and not a function of time, $t$ . Thus, we cannot simply take the integral over $t$ and must instead “change variables” to take the integral over $x$ . $x$ and $t$ are related through velocity:

\begin{align*} v &= \frac{dx}{dt}\\ \therefore dt &= \frac{1}{v}dx \end{align*}

(23)

We can thus write:

\begin{align*} dv &= a(x)dt = a(x)\frac{1}{v}dx \\ \end{align*}

(24)

The equation above is called a “separable differential equation”, which can also be written:

\frac{dv}{dx}=\frac{1}{v}a(x)

(25)

This is called a differential equation because it relates the derivative of a function (the derivative of $v$ with respect to $x$ , on the left) to the function itself ( $v$ appears on the right as well). The differential equation is “separable”, because we can separate out all of the quantities that depend on $v$ and on $x$ on different sides of the equation:

vdv = a(x)dx

(26)

This last equation says that $vdv$ is equal to $a(x)dx$ . Remember that $dx$ is the length of a very small segment in $x$ , and that $dv$ is the change in velocity over that very small segment. Since the terms on the left and right are equal, if we sum (integrate) the quantity $vdv$ over many segments, that sum must be equal to the sum (integral) of the quantity $a(x)dx$ over the same segments. Let us choose those segment such that for the beginning of the first interval the position and speed are $x_0$ and $v_0$ , respectively, and the position and speed at the end of the last segment are $X$ and $V$ , respectively. We then must have that:

\begin{align*} \int_{v_0}^{V}vdv&=\int_{x_0}^{X}a(x)dx\\ \frac{1}{2}V^2 - \frac{1}{2}v_0^2 &= \int_{x_0}^{X}a(x)dx\\ \therefore V^2 &= v_0^2 + 2\int_{x_0}^{X}a(x)dx\\ \end{align*}

(27)

which is the same as we found earlier. If the acceleration is constant, we recover our formula from kinematics:

\begin{align*} V^2 &= v_0^2+ 2\int_{x_0}^{X}adx\\ &=v_0^2+ 2a(X-x_0)\\ \therefore V^2- v_0^2 &= 2a(X-x_0) \end{align*}

(28)

Example 6.3

Figure 9:A block is launched along a frictionless surface by compressing a spring by a distance $D$ . The top panel shows the spring when at rest, and the bottom panel shows the spring compressed by a distance $D$ just before releasing the block.

A block of mass $m$ can slide freely along a frictionless surface. A horizontal spring, with spring constant, $k$ , is attached to a wall on one end, while the other end can move freely, as shown in Figure 9. A coordinate system is defined such that the $x$ axis is horizontal and the free end of the spring is at $x=0$ when the spring is at rest. The block is pushed against the spring so that the spring is compressed by a distance $D$ . The block is then released. What speed will the block have when it leaves the spring?

Solution

As you recall, the force exerted by a spring depends on the compression or extension of the spring and is given by Hooke’s Law:

\vec F(x) = -kx\hat x

(29)

where $x$ is the position of the free end of the spring and $x=0$ corresponds to the spring being at rest. In our case, when the edge of the block is located at $x_0=-D$ (the spring is compressed), the force is thus in the positive $x$ direction (since $x_0$ is a negative number).

The forces on the block are:

$\vec F_g$ , its weight, with magnitude $mg$ .
$\vec N$ , a normal force exerted by the ground.
$\vec F(x)$ , the spring force.

Since the block is not moving vertically, the magnitude of the normal force must equal the weight $N=mg$ , since these are the only forces with components in the vertical direction. The $x$ component of Newton’s Second Law gives us the acceleration of the block (which depends on $x$ ):

\begin{align*} \sum F_x = -kx &= ma(x)\\ \therefore a(x)&=-\frac{k}{m}x \end{align*}

(30)

Again, recall that if $x$ is negative, then the acceleration will be in the positive direction. Since this scenario is exactly the same that we described above in the text, namely a force that varies continuously with position, we can apply the formula that we found earlier for determining the velocity after a varying force has been applied from position $x=x_0$ to position $x=X$ :

\begin{align*} V^2 &= v_0^2 + 2\int_{x_0}^{X}a(x)dx \end{align*}

(31)

$V$ is the final speed that we would like to find, $v_0=0$ because the block starts at rest, and $x_0=-D$ is the starting position of the block. $X$ is the position along the $x$ axis where the block leaves the spring.

We have to think a little about what the value of $X$ should be: when the spring is compressed and the block accelerating, the spring is pushing the block in the positive $x$ direction. Once the block reaches $x=0$ the spring would want to pull the block backwards, but since it is not attached to the block, it stops exerting a force on the block at that point. The block thus leaves the spring at $x=0$ , so that the final position is $X=0$ . The speed of the block when it leaves the spring is thus:

\begin{align*} V^2 &= v_0^2 + 2\int_{x_0}^{X}a(x)dx\\ &= 0 + 2\int_{-D}^{0}a(x)dx\\ &= 2\int_{-D}^{0}-\frac{k}{m}xdx\\ &= 2\left[ - \frac{k}{m}\frac{1}{2}x^2\right]_{-D}^{0}\\ &= \frac{k}{m}D^2\\ \therefore V &= \sqrt{\frac{k}{m}}D \end{align*}

(32)

Discussion: This model for the speed of the block when it leaves the spring makes sense because:

The dimension for the expression for $V$ is correct (you should check this!).
If the spring is compressed more (bigger value of $D$ ), then the speed will be higher.
If the mass is bigger (more inertia), then the final speed will be lower.
If the spring is stiffer (bigger value of $k$ ), then the final speed will be higher.

If you have studied physics before, you may have realized that the speed is easily found by conservation of energy:

\frac{1}{2}mV^2=\frac{1}{2}kD^2

(33)

which gives the same value for $V$ . As we will see in a later chapter, kinetic and potential energy are defined as they are, precisely because it makes using conservation of energy equivalent to using forces as we just did.

Example 6.4

An object of mass $m$ is released from rest out of a helicopter. The drag (air-resistance) on the object can be modelled as having a magnitude given by $bv$ , where $v$ is the speed of the object and $b$ is a constant of proportionality. How does the velocity of the object depend on time?

Solution

As the object falls through the air, the forces exerted on the object are:

$F_g$ , its weight, with magnitude $mg$ , exerted downwards.
$F_d$ , the force of drag, with magnitude $bv$ , exerted upwards.

Since the object will fall in a straight line, this is a one-dimensional problem, and we can choose the $x$ axis to be vertical, with positive $x$ pointing downwards, and the origin located where the object was released. The object will thus have a positive acceleration and move in the positive $x$ direction with this choice of coordinate system. This is illustrated in the free-body diagram in Figure 10.

Figure 10:Free-body diagram for a block free-falling with drag.

Newton’s Second Law for the object gives:

\begin{align*} \sum F_x = F_g - F_d &= ma\\ mg - bv &= ma\\ \therefore a &= g-\frac{b}{m}v \end{align*}

(34)

In this case, the acceleration depends explicitly on velocity rather than position, as we had before. However, we can use the same methodology to find how the velocity changes with time. First, we can note that the acceleration is zero if:

\begin{align*} g-\frac{b}{m}v &=0\\ \therefore v = \frac{mg}{b} \end{align*}

(35)

That is, once the object reaches a speed of $v_{term}=mg/b$ , it will stop accelerating, i.e. it will reach “terminal velocity”. Note that this is the same condition as requiring that the drag force ( $bv$ ) have the same magnitude as the weight ( $mg$ ).

Writing the acceleration as $a=\frac{dv}{dt}$ , we can write:

\begin{align*} \frac{dv}{dt} &= \left(g-\frac{b}{m}v \right) \end{align*}

(36)

which again, is a separable differential equation, in which we can write the terms that depend on $v$ and those that depend on $t$ on separate sides of the equal sign:

\begin{align*} \frac{dv}{g-\frac{b}{m}v}&= dt\\ \frac{dv}{v-\frac{mg}{b}}&= -\frac{b}{m}dt\\ \end{align*}

(37)

where we re-arranged the equation in the second line so that it would be easier to integrate in the next step. We can find the velocity, $v(t)$ , at some time, $t$ , by stating that $v=0$ at $t=0$ and taking the integrals (sum) on both sides. Again, we are modelling the motion as being made up of a large number of very small segments where the quantities on both sides of the equation are the same. Thus, if we sum (integrate) those quantities over all of the same segments, the left and right hand side of the equations will still be equal to each other:

\begin{align*} \int_0^{v(t)}\frac{dv}{v-\frac{mg}{b}} &= -\int_0^t\frac{b}{m} dt\\ \left[\ln\left(v-\frac{mg}{b} \right)\right]_0^{v(t)} &=-\frac{b}{m}t\\ \ln\left(v(t)-\frac{mg}{b} \right)-\ln\left(-\frac{mg}{b} \right)&=-\frac{b}{m}t\\ \ln\left( \frac{v(t)-\frac{mg}{b}}{-\frac{mg}{b}} \right)&=-\frac{b}{m}t\\ \end{align*}

(38)

where, in the last line, we used the property that $\ln(a)-\ln(b)=\ln(a/b)$ . By taking the exponential on either side of the equation ( $e^{\ln(x)}=x$ ), we can find an expression for the velocity as a function of time:

\begin{align*} \frac{v(t)-\frac{mg}{b}}{-\frac{mg}{b}} &= e^{-\frac{b}{m}t}\\ v(t)-\frac{mg}{b} &= -\frac{mg}{b}e^{-\frac{b}{m}t}\\ \therefore v(t) &= \frac{mg}{b}-\frac{mg}{b}e^{-\frac{b}{m}t}\\ &=\frac{mg}{b}\left(1-e^{-\frac{b}{m}t}\right) \end{align*}

(39)

Discussion: This equation tells us that the velocity increases as a function of time, but the rate of increase decreases exponentially with time. At time $t=0$ , the velocity is zero, as expected. As $t$ approaches infinity, $v$ approaches, $\frac{mg}{b}$ , which is the terminal velocity. The time dependence of the velocity is illustrated in Figure 11.

$Velocity as a function of time for an object of mass $m={\rm 10{kg}}$ which is free-falling from rest with a drag coefficient $b=0.5 {\rm Ns/m}$.}$

Figure 11:Velocity as a function of time for an object of mass $m={\rm 10{kg}}$ which is free-falling from rest with a drag coefficient $b={\rm{0.5}{Ns/m}}$ .

6.4Uniform circular motion¶

As we saw in Chapter 4.1, “uniform circular motion” is defined to be motion along a circle with constant speed. This may be a good time to review Section 4.5 for the kinematics of motion along a circle. In particular, for the uniform circular motion of an object around a circle of radius $R$ , you should recall that:

The velocity vector, $\vec v$ , is always tangent to the circle.
The acceleration vector, $\vec a$ , is always perpendicular to the velocity vector, because the magnitude of the velocity vector does not change.
The acceleration vector, $\vec a$ , always points towards the centre of the circle.
The acceleration vector has magnitude $a=v^2/R$ .
The angular velocity, ω, is related to the magnitude of the velocity vector by $v=\omega R$ and is constant.
The angular acceleration, α, is zero for uniform circular motion, since the angular velocity does not change.

In particular, you should recall that even if the speed is constant, the acceleration vector is always non-zero in uniform circular motion because the velocity changes direction. According to Newton’s Second Law, this implies that there must be a net force on the object that is directed towards the centre of the circle^[1] (parallel to the acceleration):

\sum \vec F = m\vec a

(40)

where the acceleration has a magnitude $a=v^2/R$ . Because the acceleration is directed towards the centre of the circle, we sometimes call it a “radial” acceleration (parallel to the radius), $a_R$ , or a “centripetal” acceleration (directed towards the centre), $a_c$ .

Consider an object in uniform circular motion in a horizontal plane on a frictionless surface, as depicted in Figure 12.

Figure 12:An object undergoing uniform circular motion on a frictionless surface, as seen from above.

The only way for the object to undergo uniform circular motion as depicted is if the net force on the object is directed towards the centre of the circle. One way to have a force that is directed towards the centre of the circle is to attach a string between the center of the circle and the object, as shown in Figure 12. If the string is under tension, the force of tension will always be towards the centre of the circle. The forces on the object are thus:

$\vec F_g$ , its weight with magnitude $mg$ .
$\vec N$ , a normal forced exerted by the surface.
$\vec T$ , a force of tension exerted by the string.

The forces are depicted in the free-body diagram shown in Figure 13 (as viewed from the side), where we also drew the acceleration vector. Note that this free-body diagram is only “valid” at a particular instant in time since the acceleration vector continuously changes direction and would not always be lined up with the $x$ axis.

(a)Free-body diagram (side view) for the object from [](#fig:applyingnewtonslaws:circleH_fbd)

The $y$ component just tells us that the normal force must have the same magnitude as the weight because the object is not accelerating in the vertical direction. The $x$ component tells us the relation between the magnitudes of the tension in the string and the radial acceleration. Using the speed of the object, we can also write the relation between the tension and the speed:

\begin{align*} T &= ma_R=m\frac{v^2}{R}\\ \end{align*}

(41)

Thus, we find that the tension in the string increases with the square of the speed, and decreases with the radius of the circle.

Example 6.5

Figure 15:A car going around a curve that can be approximated as the arc of a circle of radius $R$ .

A car goes around a curve which can be approximated as the arc of a circle of radius $R$ , as shown in Figure 15. The coefficient of static friction between the tires of the car and the road is $\mu_s$ . What is the maximum speed with which the car can go around the curve without skidding?

Solution

If the car is going at constant speed around a circle, then the sum of the forces on the car must be directed towards the centre of the circle. The only force on the car that could be directed towards the centre of the circle is the force of friction between the tires and the road. If the road were perfectly slick (think driving in icy conditions), it would not be possible to drive around a curve since there could be no force of friction. The forces on the car are:

$\vec F_g$ , its weight with magnitude $mg$ .
$\vec N$ , a normal force exerted upwards by the road.
$\vec f_s$ , a force of static friction between the tires and the road. This is static friction, because the surface of the tire does not move relative to the surface of the road if the car is not skidding. The force of static friction has a magnitude that is at most $f_s\leq\mu_sN$ .

The forces on the car are shown in the free-body diagram in Figure 16.

Figure 16:Free-body diagram for the car as seen looking at the car from the back (the centre of the curve is towards the left).

The $y$ component of Newton’s Second Law tells us that the normal force exerted by the road must equal the weight of the car:

\begin{align*} \sum F_y = N-F_g&=0\\ \therefore N &=mg \end{align*}

(42)

The $x$ component relates the force of friction to the radial acceleration (and thus to the speed):

\begin{align*} \sum F_x = f_s =ma_R&=m\frac{v^2}{R}\\ \therefore f_s &= m\frac{v^2}{R} \end{align*}

(43)

The force of friction must be less than or equal to $f_s\leq\mu_sN=\mu_smg$ (since $N=mg$ from the $y$ component of Newton’s Second Law), which gives us a condition on the speed:

\begin{align*} f_s = m\frac{v^2}{R}&\leq\mu_smg\\ v^2 &\leq \mu_s g R\\ \therefore v &\leq \sqrt{\mu_s g R} \end{align*}

(44)

Thus, if the speed is less than $\sqrt{\mu_s g R}$ , the car will not skid and the magnitude of the force of static friction, which results in an acceleration towards the centre of the circle, will be smaller or equal to its maximal possible value.

Discussion: The model for the maximum speed that the car can travel around the curve makes sense because:

The dimension of $\sqrt{\mu_s g R}$ is speed.
The speed is larger if the radius of the curve is larger (one can go faster around a wider curve without skidding).
The speed is larger if the coefficient of friction is large (if the force of friction is larger, a larger radial acceleration can be sustained).

Example 6.6

Figure 17:A ball attached to a string undergoing circular motion in a vertical plane.

A ball is attached to a mass-less string and executing circular motion along a circle of radius $R$ that is in the vertical plane, as depicted in Figure 17. Can the speed of the ball be constant? What is the minimum speed of the ball at the top of the circle if it is able to make it around the circle?

Solution

The forces that are acting on the ball are:

$\vec F_g$ , its weight with magnitude $mg$ .
$\vec T$ , a force of tension exerted by the string.

Figure 18 shows the free-body diagram for the forces on the ball at three different locations along the path of the circle.

Figure 18:A ball attached to a string undergoing circular motion in a vertical plane.

In order for the ball to go around in a circle, there must be at least a component of the net force on the ball that is directed towards the centre of the circle at all times. In the bottom half of the circle (positions 1 and 2), only the tension can have a component directed towards the centre of the circle.

Consider in particular the position labelled 2, when the string is horizontal and the tension is equal to $\vec T_2$ . The free-body diagram in Figure 18 also shows the vector sum of the weight and tension at position 2 (the red arrow labelled $\sum \vec F$ ), which points downwards and to the left. It is thus clearly impossible for the acceleration vector to point towards the centre of the circle, and the acceleration will have components that are both tangential ( $a_T$ ) to the circle and radial ( $a_R$ ), as shown by the vector $\vec a_2$ in Figure 18.

The radial component of the acceleration will change the direction of the velocity vector so that the ball remains on the circle, and the tangential component will reduce the magnitude of the velocity vector. According to our model, it is thus impossible for the ball to go around the circle at constant speed, and the speed must decrease as it goes from position 2 to position 3, no matter how one pulls on the string (you can convince yourself of this by drawing the free-body diagram at any point between points 2 and 3).

The minimum speed for the ball at the top of the circle is given by the condition that the tension in the string is zero just at the top of the trajectory (position 3). The ball can still go around the circle because, at position 3, gravity is towards the centre of the circle and can thus give an acceleration that is radial, even with no tension. The $y$ component of Newton’s Second Law, at position 3 gives:

\begin{align*} \sum F_y = -F_g &= ma_y\\ \therefore a_y &=-g \end{align*}

(45)

The magnitude of the acceleration is the radial acceleration, and is thus related to the speed at the top of the trajectory:

\begin{align*} a_R&=-a_y=g = \frac{v^2}{R}\\ \therefore v_{min}&=\sqrt{gR} \end{align*}

(46)

which is the minimum speed at the top of the trajectory for the ball to be able to continue along the circle. The tension in the string would change as the ball moves around the circle, and will be highest at the bottom of the trajectory, since the tension has to be bigger than gravity so that the net force at the bottom of the trajectory is upwards (towards the centre of the circle).

Discussion: The model for the minimum speed of the ball at the top of the circle makes sense because:

$\sqrt{gR}$ has the dimension of speed.
The minimum velocity is larger if the circle has a larger radius (try this with a mass attached at the end of a string).
The minimum velocity is larger if the mass is bigger (again, try this at home!).

6.4.1Banked curves¶

As we saw in Example 6.5, there is a maximum speed with which a car can go around a curve before it starts to skid. You may have noticed that roads, highways especially, are banked where there are curves. Racetracks for cars that go around an oval (the boring kind of car races) also have banked curves. As we will see, this allows the speed of vehicles to be higher when going around the curve; or rather, it makes the curves safer as the speed at which vehicles would skid is higher. In Example 6.5, we saw that it was the force of static friction between the tires of the car and the road that provided the only force with a component towards the centre of the circle. The idea of using a banked curve is to change the direction of the normal force between the road and the car tires so that it, too, has a component in the direction towards the centre of the circle.

Consider the car depicted in Figure 19 which is seen from behind making a left turn around a curve that is banked by an angle θ with respect to the horizontal and can be modelled as an arc from a circle of radius $R$ .

Figure 19:A car moving into the page and going around a banked curved so that it is turning towards the left (the centre of the circle is to the left).

The forces exerted on the car are the same as in Example 6.5, except that they point in different directions. The forces are:

$\vec F_g$ , its weight with magnitude $mg$ .
$\vec N$ , a normal force exerted by the road, perpendicular to the surface of the road.
$\vec f_s$ , a force of static friction between the tires and the road. This is static friction, because the surface of the tire does not move relative to the surface of the road if the car is not skidding. The force of static friction has a magnitude that is at most $f_s\leq\mu_sN$ and is perpendicular to the normal force. The force could be either upwards or downwards, depending on the other forces on the car.

A free-body diagram for the forces on the car is shown in Figure 20, along with the acceleration (which is in the radial direction, towards the centre of the circle), and our choice of coordinate system (choosing $x$ parallel to the acceleration). The direction of the force of static friction is not known a priori and depends on the speed of the car:

If the speed of the car is zero, the force of static friction is upwards. With a speed of zero, the radial acceleration is zero, and the sum of the forces must thus be zero. The impeding motion of the car would be to slide down the banked curve (just like a block on an incline).
If the speed of the car is very large, the force of static friction is downwards, as the impeding motion of the car would be to slide up the bank. The natural motion of the car is to go in a straight line (Newton’s First Law). If the components of the normal force and of the force of static friction directed towards the centre of the circle are too small to allow the car to turn, then the car would slide up the bank (so the impeding motion is up the bank and the force of static friction is downwards).

Figure 20:Free-body diagram for the forces on the car. The direction of the force of static friction cannot be determined, as it depends on the acceleration of the car, so it is shown twice (with dotted lines).

There is thus an “ideal speed” at which the force of static friction is precisely zero, and the $x$ component of the normal force is responsible for the radial acceleration. At higher speeds, the force of static friction is downwards and increases in magnitude to keep the car’s acceleration towards the centre of the circle. At some maximal speed, the force of friction will reach its maximal value, and no longer be able to keep the car’s acceleration pointing towards the centre of the circle. At speeds lower than the ideal speed, the force of friction is directed upwards to prevent the car from sliding down the bank. If the coefficient of static friction is too low, it is possible that at low speeds, the car would start to slide down the bank (so there would be a minimum speed below which the car would start to slide down).

Let us model the situation where the force of static friction is identically zero so that we can determine the ideal speed for the banked curve. The only two forces on the car are thus its weight and the normal force. The $x$ and $y$ component of Newton’s Second Law give:

\begin{align*} \sum F_x &= N\sin\theta = ma_R=m\frac{v^2}{R}\nonumber\\ \therefore N\sin\theta &= m\frac{v^2}{R} \end{align*}

(47)

\begin{align*} \sum F_y &= N\cos\theta-F_g = 0\nonumber\\ \therefore N\cos\theta&=mg \end{align*}

(48)

We can divide Equation (47) by Equation (48), noting that $\tan\theta=\sin\theta/\cos\theta$ , to obtain:

\begin{align*} \tan\theta &= \frac{v^2}{gR}\\ \therefore v_{ideal} &=\sqrt{gR\tan\theta} \end{align*}

(49)

At this speed, the force of static friction is zero. In practice, one would use this equation to determine which bank angle to use when designing a road, so that the ideal speed is around the speed limit or the average speed of traffic. We leave it as an exercise to determine the maximal speed that the car can go around the curve before sliding out.

6.4.2Inertial forces in circular motion¶

As you sit in a car that is going around a curve, you will feel pushed outwards, away from the centre of the circle that the car is going around. This is because of your inertia (Newton’s First Law), and your body would go in a straight line if the car were not exerting a net force on you towards the centre of the circle. You are not so much feeling a force that is pushing you outwards as you are feeling the effects of the car seat pushing you inwards; if you were leaning against the side of the car that is on the outside of the curve, you would feel the side of the car pushing you inwards towards the centre of the curve, even if it “feels” like you are pushing outwards against the side of the car.

If we model your motion looking at you from the ground, we would include a force of friction between the car seat (or the side of the car, or both) and you that is pointing towards the centre of the circle, so that the sum of the forces exerted on you is towards the centre of the circle. We can also model your motion from the non-inertial frame of the car. As you recall, because this is a non-inertial frame of reference, we need to include an additional inertial force, $\vec F_I$ , that points opposite of the acceleration of the car, with magnitude $F_I=ma_R$ (if the net acceleration of the car is $a_R$ ). Inside the non-inertial frame of reference of the car, your acceleration (relative to the reference frame, i.e. the car) is zero. This is illustrated by the diagrams in Figure 21.

Figure 21:(Left:) A person sitting on a car seat in a car turning towards the left. (Centre:) Free-body diagram for the person as modelled in the inertial reference frame of the ground. (Right:) Free-body diagram for the person as modelled in the non-inertial frame of reference of the car, including an additional inertial force.

The $y$ component of Newton’s Second Law in both frames of reference is the same:

\begin{align*} \sum F_y&=N-F_g=0\\ \therefore N&=mg \end{align*}

(50)

and simply tells us that the normal force is equal to the weight. In the reference frame of the ground, the $x$ component of Newton’s Second Law gives:

\begin{align*} \sum F_x &= f_s = ma_R\\ \therefore f_s &= m\frac{v^2}{R} \end{align*}

(51)

In the frame of reference of the car, where your acceleration is zero and an inertial force of magnitude $F_I=mv^2/R$ is exerted on you, the $x$ component of Newton’s Second Law gives:

\begin{align*} \sum F_x &= f_s-F_I = 0\\ \therefore f_s - m\frac{v^2}{R} &= 0 \end{align*}

(52)

which of course, mathematically, is exactly equivalent. The inertial force is not a real force in the sense that it is not exerted by anything. It only comes into play because we are trying to use Newton’s Laws in a non-inertial frame of reference. However, it does provide a good model for describing the sensation that we have of being pushed outwards when the car goes around a curve. Sometimes, people will refer to this force as a “centrifugal” force, which means “a force that points away from the centre”. You should however remember that this is not a real force exerted on the object, but is the result of modelling motion in a non-inertial frame of reference.

6.5Non-uniform circular motion¶

In non-uniform circular motion, an object’s motion is along a circle, but the object’s speed is not constant. In particular, the following will be true

The object’s velocity vector is always tangent to the circle.
The speed and angular speed of the object are not constant.
The angular acceleration of the object is not zero.
The acceleration vector will not point towards the centre of the circle.

Since the acceleration vector does not point towards the centre of the circle, it is usually convenient to break up the acceleration vector into two components: $a_R$ , a component that is radial (towards the centre of the circle), and $a_T$ , a component that is tangent to the circle (and perpendicular to to the radial component). The radial component is “responsible” for the change in direction of the velocity such that the object goes in a circle. the magnitude of the radial acceleration is the same as it is for uniform circular motion:

a_R=\frac{v^2}{r}

(53)

where the speed is no longer constant in time. The tangential component of the acceleration is responsible for changing the magnitude of the velocity of the object:

a_T = \frac{dv}{dt}

(54)

Example 6.7

Figure 22:An ant on a horizontal turntable that is starting to spin, as seen from above.

A small ant is sleeping on a turntable just as the turntable starts to spin from rest, with an angular acceleration $\alpha=1 {\rm rad/s^2}$ that is small enough so that, initially, the ant remains on the turntable. The ant is a distance $R=0.1 {\rm m}$ from the centre of the turntable, as shown in Figure 22 and the coefficient of static friction between the ant’s “feet” and the turntable is $\mu_s=0.5$ . After how much time will the ant slide off from the turntable?

6.6Summary¶

When the velocity of an object does not change direction continuously (“linear motion”), we can model its motion independently over several segments in such a way that the motion is one dimensional in each segment. This allows us to choose a coordinate system in each segment where the acceleration vector is co-linear with one of the axes.

When the forces on an object changes continuously, we need to use calculus to determine the motion of the object. If the velocity vector for an object changes direction continuously, we need to model the motion in each dimension independently.

If an object undergoes uniform circular motion, the acceleration vector and the sum of the forces always point towards the centre of the circle. In the radial direction, Newton’s Second Law gives

\sum \vec F = ma_R = m\frac{v^2}{R}

(64)

If an object’s speed is changing as it moves around a circle the acceleration vector will have a component that is towards the centre of the circle (the radial component) and a component that is tangential to the circle. The tangential component is responsible for the change in speed, whereas the radial component is responsible for the change in direction of the velocity.

In a reference frame that is rotating about a circle, an inertial force, sometimes called the centrifugal force, appears to push all objects co-moving with the reference frame towards the outside of the circle.

6.7Thinking about the material¶

6.8Sample problems and solutions¶

6.8.1Problems¶

6.8.2Solutions¶

Part 1 - Mechanics

Chapter 5 - Newton's Laws

Part 1 - Mechanics

Chapter 7 - Work and energy