Problem 8 - College Physics

Question

\(\bullet\) A 750 gram grinding wheel 25.0 \(\mathrm{cm}\) in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 220 \(\mathrm{rpm}\) about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 45.0 s with constant angular acceleration due to friction at the axle. What torque does friction exert while this wheel is slowing down?

Step-by-Step Solution

Answer

The torque due to friction is approximately \(-0.003\) N·m.

Step 1: Calculate the Moment of Inertia

The grinding wheel is modeled as a solid disk. The moment of inertia for a solid disk about its center is given by \( I = \frac{1}{2}mr^2 \). Here, \( m = 0.75 \) kg is the mass of the wheel, and \( r = 12.5 \) cm = 0.125 m is the radius. Thus, \( I = \frac{1}{2} \times 0.75 \times (0.125)^2 = 0.005859375 \) kg·m².

Step 2: Determine Initial and Final Angular Velocities

Angular velocity is given in revolutions per minute (rpm) but needs to be converted to radians per second (rad/s) for calculations. The initial angular velocity \( \omega_i = 220 \) rpm \( \approx 220 \times \frac{2\pi}{60} = 23.038 \) rad/s. The final angular velocity \( \omega_f = 0 \) rad/s since the wheel stops.

Step 3: Find Angular Acceleration

The angular acceleration can be found using the kinematic equation \( \omega_f = \omega_i + \alpha t \), where \( \alpha \) is the angular acceleration and \( t = 45.0 \) s. Rearranging gives \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 23.038}{45.0} = -0.511 \) rad/s².

Step 4: Calculate Torque

Torque \( \tau \) is related to angular acceleration \( \alpha \) by the equation \( \tau = I\alpha \). Substituting known values: \( \tau = 0.005859375 \times (-0.511) = -0.002995 \) N·m. The negative sign indicates that the torque acts in the opposite direction of the initial motion, i.e., it's slowing the wheel down.

Key Concepts

Moment of Inertia

The moment of inertia is a core concept in physics that describes how difficult it is to change the rotation of an object. Think of it like rotational mass. The larger the moment of inertia, the more effort it takes to spin something around. For our grinding wheel shaped like a solid disk, the formula for the moment of inertia is used: \( I = \frac{1}{2}mr^2 \).
This formula shows that the moment of inertia depends on two things:

• Mass \( (m) \)
• Radius \( (r) \)

Substituting the given values of mass (0.75 kg) and radius (0.125 m) into the formula, we get 0.005859375 kg·m². Understanding this helps us see how size and mass affect how an object spins.

Angular Acceleration

Angular acceleration is like acceleration but for rotation. It tells us how quickly something is speeding up or slowing down its spin. In the grinding wheel problem, we know that the wheel stops spinning in 45 seconds.
We can use the formula from physics, \( \omega_f = \omega_i + \alpha t \), to find angular acceleration \( \alpha \). This formula re-arranges to \( \alpha = \frac{\omega_f - \omega_i}{t} \), where:

• \( \omega_i \) is the initial angular velocity
• \( \omega_f \) is the final angular velocity (which is 0 when the wheel stops)
• \( t \) is time

For our exercise, the initial velocity \( \omega_i \) is 23.038 rad/s, and the final \( \omega_f \) is 0 rad/s. By plugging these into our rearranged formula, we find that \( \alpha \) is -0.511 rad/s². The negative sign tells us that the wheel is slowing down, not speeding up.

Kinematic Equations

Kinematic equations are essential in connecting different physical quantities in motion. These equations help us understand the change in angular position and helping to calculate angular acceleration like we did.
For rotational motion, a key kinematic equation is \( \omega_f = \omega_i + \alpha t \). Knowing this helps when there’s a change in speed over some time. It gives the relationship between the initial & final velocities and the acceleration.
This equation greatly simplifies problems involving spinning objects because:

• You can determine how long it takes to stop or start something.
• You can understand how factors like torque influence motion.

By mastering this concept, you can solve many physics problems involving rotation and Angular Kinematics.

Angular Velocity Conversion

Angular velocity is typically measured in either revolutions per minute (rpm) or radians per second (rad/s). But for calculations, rad/s is more useful because it connects directly with the formulas we use. That’s why converting rpm to rad/s is often one of the first steps.

• To convert rpm to rad/s, the formula is \( \text{rad/s} = \text{rpm} \times \frac{2\pi}{60} \).

For our grinding wheel exercise, the wheel spins at 220 rpm, and using the formula, this converts to approximately 23.038 rad/s. This step is crucial because it properly sets up the values for solving other parts of the problem like angular acceleration.