8.2.1 Dynamics of Uniform Circular Motion

A photograph of a sports car on a banked track. — On banked curves, the normal force of the road assists in providing the centripetal acceleration of the turn.

A particle in uniform circular motion is clearly not traveling at constant velocity in a straight line. Consequently, according to Newton’s first law, the particle must have a net force acting on it. We’ve already determined the acceleration of a particle in uniform circular motion—the centripetal acceleration of Equation 8.4. Newton’s second law tells us exactly how much net force is needed to cause this acceleration:

{\vec{F}}_{net} = m \vec{a} = (\frac{m v^{2}}{r}, toward center of circle)

(8.6)

In other words, a particle of mass m moving at constant speed v around a circle of radius r must have a net force of magnitude $m v^{2} / r$ pointing toward the center of the circle. Without such a force, the particle would move off in a straight line tangent to the circle.

FIGURE 8.4 shows the net force $\vec{F}_{net}$ acting on a particle as it undergoes uniform circular motion. You can see that $\vec{F}_{n e t}$ , like $\vec{a},$ points along the radial axis of the rtz-coordinate system, toward the center of the circle. The tangential and perpendicular components of $\vec{F}_{net}$ are zero.

The figure illustrates the net force acting on a particle in a uniform circular motion.

FIGURE 8.4 Full Alternative Text

NOTE The force described by Equation 8.6 is not a new force. The force itself must have an identifiable agent and will be one of our familiar forces, such as tension, friction, or the normal force. Equation 8.6 simply tells us how the force needs to act—how strongly and in which direction—to cause the particle to move with speed v in a circle of radius r.

The usefulness of the rtz-coordinate system becomes apparent when we write Newton’s second law, Equation 8.6, in terms of the r-, t-, and z-components:

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

(8.7)

For uniform circular motion, the sum of the forces along the t-axis and along the z-axis must equal zero, and the sum of the forces along the r-axis must equal $m a_{r},$ where $a_{r}$ is the centripetal acceleration.

EXAMPLE 8.2 Spinning in a circle

An energetic father places his 20 kg child on a 5.0 kg cart to which a 2.0-m-long rope is attached. He then holds the end of the rope and spins the cart and child around in a circle, keeping the rope parallel to the ground. If the tension in the rope is 100 N, how many revolutions per minute (rpm) does the cart make? Rolling friction between the cart’s wheels and the ground is negligible.

MODEL Model the child in the cart as a particle in uniform circular motion.

VISUALIZE FIGURE 8.5 shows the pictorial representation. A circular-motion problem usually does not have starting and ending points like a projectile problem, so numerical subscripts such as $x_{1} or y_{2}$ are usually not needed. Here we need to define the cart’s speed $v$ and the radius r of the circle. Further, a motion diagram is not needed for uniform circular motion because we already know the acceleration $\vec{a}$ points to the center of the circle.

The figure consists of two parts that illustrate a pictorial representation and a free-body diagram of a cart spinning in a circle.

FIGURE 8.5 Full Alternative Text

The essential part of the pictorial representation is the free-body diagram. For uniform circular motion we’ll draw the free-body diagram in the rz-plane, looking at the edge of the circle, because this is the plane of the forces. The contact forces acting on the cart are the normal force of the ground and the tension force of the rope. The normal force is perpendicular to the plane of the motion and thus in the z-direction. The direction of $\vec{T}$ is determined by the statement that the rope is parallel to the ground. In addition, there is the long-range gravitational force $\vec{F}_{G}$ .

SOLVE We defined the r-axis to point toward the center of the circle, so $\vec{T}$ points in the positive r-direction and has r-component $T_{r} = T$ . Newton’s second law, using the rtz-components of Equations 8.7, is

\begin{array}{l} \sum^{​} F_{r} = T = \frac{m v^{2}}{r} \\ \sum^{​} F_{z} = n - m g = 0 \end{array}

We’ve taken the r- and z-components of the forces directly from the free-body diagram, as you learned to do in Chapter 6. Then we’ve explicitly equated the sums to $a_{r} = v^{2} / r$ and $a_{z} = 0$ . This is the basic strategy for all uniform circular-motion problems. From the z-equation we can find that $n = m g$ . This would be useful if we needed to determine a friction force, but it’s not needed in this problem. From the r-equation, the speed of the cart is

v = \sqrt{\frac{r T}{m}} = \sqrt{\frac{(2.0 m) (100 N)}{25 kg}} = 2.83 m /s

The cart’s angular velocity is then found from Equations 8.5:

ω = \frac{v_{t}}{r} = \frac{v}{r} = \frac{2.83 m / s}{2 .0 m} = 1.41 rad / s

This is another case where we inserted the radian unit because $ω$ is specifically an angular velocity. Finally, we need to convert $ω$ to rpm:

ω = \frac{1.41 rad}{1 s} \times \frac{1 rev}{2 π rad} \times \frac{60 s}{1 min} = 14 rpm

REVIEW 14 rpm corresponds to a period $T \approx 4 s$ . This result is reasonable.

8.2.1 Dynamics of Uniform Circular Motion

FIGURE 8.4 The net force points in the radial direction, toward the center of the circle.

EXAMPLE 8.2 Spinning in a circle

FIGURE 8.5 Pictorial representation of a cart spinning in a circle.