8.5 Nonuniform Circular Motion

Many interesting examples of circular motion involve objects whose speed changes. As we’ve already noted, a roller-coaster car doing a loop-the-loop slows down as it goes up one side, speeds up as it comes back down the other side. Circular motion with a changing speed is called nonuniform circular motion.

FIGURE 8.16 shows a particle moving in a circle of radius r. In addition to a radial force component—required for all circular motion—this particle experiences a tangential force component $(F_{net})_{t}$ and hence a tangential acceleration

a_{t} = \frac{d v_{t}}{d t}

(8.21)

Now $v_{t}$ is the particle’s velocity around the circle, with speed $v = | v_{t} |$ , so a tangential force component causes the particle to change speed. That is, the particle is undergoing nonuniform circular motion. Note that $(F_{net})_{t}$ , like $v_{t}$ , is positive when ccw, negative when cw.

The figure illustrates the radial and tangential components of net force applied to a particle in circular motion.

FIGURE 8.16 Full Alternative Text

Force and acceleration are still related to each other through Newton’s second law:

\begin{array}{l} {(F_{net})}_{r} = \sum^{​} F_{r} = m a_{r} = \frac{m v_{t} ​^{2}}{r} = m ω^{2} r \\ {(F_{net})}_{t} = \sum^{​} F_{t} = m a_{t} \\ {(F_{net})}_{z} = \sum^{​} F_{z} = 0 \end{array}

(8.22)

If the tangential force is constant, you can apply what you know about constant-acceleration kinematics to solve for $v_{t}$ at a later time.

NOTE Equations 8.22 differ from Equations 8.7 for uniform circular motion only in the fact that $a_{t}$ is no longer zero.

EXAMPLE 8.6 Sliding out of the curve

A 1500 kg car drives around a flat, 50-m-diameter track, starting from rest. The drive wheels supply a small but steady 525 N force in the forward direction. The coefficient of static friction between the car tires and the road is 0.90. How many revolutions of the track have been made when the car slides out of the curve?

MODEL Model the car as a particle in nonuniform circular motion. Assume that rolling friction and air resistance can be neglected.

VISUALIZE FIGURE 8.17 shows a pictorial representation. As in earlier examples, it’s static friction, perpendicular to the tires, that causes the centripetal acceleration of circular motion. The propulsion force is a tangential force. For the first time, we need a free-body diagram showing forces in three dimensions.

The figure illustrates an object in circular motion and its free-body diagram.

FIGURE 8.17 Full Alternative Text

SOLVE At slow speeds, static friction in the radial direction keeps the car moving in a circle. But there’s an upper limit to the size of the static friction force, and the car will begin to slide out of the curve when that limit is reached. The r-component of Newton’s second law is

\sum^{​} F_{r} = f_{s} = \frac{m v_{t}^{2}}{r}

That is, static friction increases proportional to $v_{t}^{2}$ until the car reaches a velocity $v_{max}$ at which the static friction is $f_{s max}$ .

Recall that the maximum possible static friction is $f_{s max} = μ_{s} n$ . We can find the normal force from the z-component of Newton’s second law:

\sum F_{z} = n - F_{G} = 0

Thus $n = F_{G} = m g$ and $f_{s max} = μ_{s} m g$ . Combining these two equations, we see that the mass cancels and we have

v_{max}^{2} = μ_{s} r g

How far does the car have to travel to reach this speed? We can find the car’s tangential acceleration from the t-component of the second law: $a_{t} = F_{wheels} / m$ . This is a constant acceleration, so we can use constant-acceleration kinematics. Let s measure the distance around the circle—the arc length. Thus, because the initial velocity is $v_{0} = 0$ , we have

v_{t}^{2} = v_{0}^{2} + 2 a_{s} s = 2 a_{s} s = \frac{2 s F_{wheels}}{m}

You’ll recall that the angular displacement, measured in radians, is $Δ θ = s / r$ . We can see from the above equation that the car travels distance $s = m v_{t}^{2} / 2 F_{wheels}$ to reach velocity $v_{t}$ , and at this time it has revolved through an angle

Δ θ = \frac{s}{r} = \frac{m v_{t}^{2}}{2 r F_{wheels}}

Using the maximum speed before sliding, we find that the car slides out of the curve after revolving through an angle

Δ θ_{max} = \frac{m v_{max}^{2}}{2 r F_{wheels}} = \frac{m}{2 r F_{wheels}} \times μ_{s} r g = \frac{μ_{s} m g}{2 F_{wheels}}

For the car in this problem,

\begin{array}{l} Δ θ_{max} & = & \frac{(0.90) (1500 kg)(9.80 m / s^{2})}{2 (525 N)} \\ = & 12.6 rad \times \frac{1 rev}{2 π rad} = 2.0 rev \end{array}

It completes 2.0 revolutions before it starts to slide.

REVIEW A 525 N force on a 1500 kg car causes a tangential acceleration of $a_{t} \approx 0.3 m / s^{2}$ . That’s a quite modest acceleration, so it seems reasonable that the car would complete 2 rev before gaining enough speed to start sliding.

We’ve come a long way since our first dynamics problems in Chapter 6, but our basic strategy has not changed.

PROBLEM-SOLVING STRATEGY 8.1
Circular-motion problems

MODEL Model the object as a particle and make other simplifying assumptions.

VISUALIZE Draw a pictorial representation. Use rtz-coordinates.

Establish a coordinate system with the r-axis pointing toward the center of the circle.
Show important points in the motion on a sketch. Define symbols and identify what the problem is trying to find.
Identify the forces and show them on a free-body diagram.

SOLVE Newton’s second law is

\begin{matrix} (F_{net})_{r} & = & \sum F_{r} = m a_{r} = \frac{m v_{t}^{2}}{r} = m ω^{2} r \\ (F_{net})_{t} & = & \sum F_{t} = m a_{t} \\ (F_{net})_{z} & = & \sum F_{z} = 0 \end{matrix}

Determine the force components from the free-body diagram. Be careful with signs.
The tangential acceleration for uniform circular motion is $a_{t} = 0$ .
Solve for the acceleration, then use kinematics to find velocities and positions.

REVIEW Check that your result has the correct units and significant figures, is reasonable, and answers the question.

Workbook Exercise 11

CHAPTER 8 CHALLENGE EXAMPLE Swinging on two strings

The 250 g ball shown in FIGURE 8.18 revolves in a horizontal plane as the vertical shaft spins. What is the critical angular speed, in rpm, that the shaft must exceed to keep both strings taut?

The figure illustrates the revolution of a ball around a vertical shaft.

FIGURE 8.18 Full Alternative Text

MODEL Model the ball as a particle in uniform circular motion. For both strings to be straight, as shown, both must be under tension. If the angular speed is slowly decreased, eventually the lower string will go slack and the ball will sag. The critical angular speed $ω_{c}$ is the angular speed at which the tension in the lower string reaches zero. We need to find an expression for the tension in the lower string, then determine when that tension becomes zero.

VISUALIZE FIGURE 8.19 is the ball’s free-body diagram with the r-axis pointing toward the center of the circle. The ball is acted on by two tension forces, at equal angles above and below horizontal, and by gravity. The free-body diagram is similar to Example 8.5, the rock in the sling, but with an additional tension force.

The figure illustrates a free-body diagram of a ball attached to a shaft using two strings.

FIGURE 8.19 Full Alternative Text

SOLVE This is uniform circular motion, so we need to consider only the r- and z-components of Newton’s second law. All the information is on the free-body diagram, where we see that gravity has only a z-component but the tensions have both r- and z-components. The two equations are

\begin{array}{l} \sum^{​} F_{r} = T_{top} cos θ + T_{bot} cos θ = m ω^{2} r \\ \sum^{​} F_{z} = T_{top} sin θ - T_{bot} sin θ - m g = 0 \end{array}

Factoring out the $cos θ$ and $sin θ$ terms, we have two simultaneous equations:

\begin{matrix} T_{top} + T_{bot} = \frac{m ω^{2} r}{cos θ} \\ T_{top} - T_{bot} = \frac{m g}{sin θ} \end{matrix}

Subtracting the second equation from the first will eliminate $T_{top} :$

2 T_{bot} = \frac{m ω^{2} r}{cos θ} - \frac{m g}{sin θ}

and thus

T_{bot} = \frac{m}{2} (\frac{ω^{2} r}{cos θ} - \frac{g}{sin θ})

You can see that this expression becomes negative—a physically impossible situation—if $ω$ is too small. The angular speed at which the tension reaches zero—the critical angular speed—is found by setting the expression in parentheses equal to zero. This gives

ω_{c} = \sqrt{\frac{g}{r tan θ}}

For our situation,

\begin{matrix} r & = & \sqrt{(1.0 {m)}^{2} - (0.50 {m)}^{2}} = 0.866 m \\ θ & = & {sin}^{- 1} [(0.50 m)/ (1.0 m)] = 30 ° \end{matrix}

Thus the critical angular speed is

ω_{c} = \sqrt{\frac{9.80 {m/s}^{2}}{(0.866 m)tan 30 °}} = 4.40 rad / s

Converting to rpm:

ω_{c} = 4.40 rad / s \times \frac{1 rev}{2 π rad} \times \frac{60 s}{1 min} = 42 rpm

REVIEW $ω_{c}$ is the minimum angular speed needed to keep both strings taut. For a ball attached to meter-long strings, 42 rpm—a bit less than 1 revolution per second—seems plausible. Your intuition probably suggests that the bottom string wouldn’t be taut if the shaft spun at only a few rpm, and hundreds of rpm seems much too high. Remember that the goal of Review is not to prove that an answer is correct but to rule out answers that, with a little thought, are clearly wrong.

8.5 Nonuniform Circular Motion

FIGURE 8.16 Net force $\vec{F}_{net}$ is applied to a particle moving in a circle.

EXAMPLE 8.6 Sliding out of the curve

FIGURE 8.17 Pictorial representation of a car speeding up around a circle.

PROBLEM-SOLVING STRATEGY 8.1
Circular-motion problems

CHAPTER 8 CHALLENGE EXAMPLE Swinging on two strings

FIGURE 8.18 A ball revolving on two strings.

FIGURE 8.19 Free-body diagram of the ball.