Question

A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3.00 cm. It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by 22.5 rad/s ?

Short answer

Angular acceleration is 66.67 rad/s²; time to decrease speed is 0.337 s.

Step-by-step solution

Determine the Moment of Inertia

To find the angular acceleration, we first need to find the moment of inertia of the sphere. The formula for the moment of inertia \( I \) of a solid sphere about an axle through its center is \( I = \frac{2}{5}mr^2 \), where \( m \) is the mass and \( r \) is the radius. The mass \( m \) is 225 g or 0.225 kg and the diameter is 3.00 cm, so the radius \( r \) is 1.50 cm or 0.015 m.Plug these numbers into the formula:\[ I = \frac{2}{5}(0.225)(0.015)^2 \]Calculate the value:\[ I = \frac{2}{5}(0.225)(0.000225) = 4.5 \times 10^{-6} \text{ kg} \cdot \text{m}^2 \]

Find Angular Acceleration

Use Newton’s second law for rotation, \( \tau = I\alpha \), where \( \tau \) is the torque and \( \alpha \) is the angular acceleration. The torque \( \tau \) can be calculated by the friction force \( f \) and the radius \( r \):\( \tau = fr = 0.0200 \times 0.015 \).Calculate the torque:\[ \tau = 3 \times 10^{-4} \text{ N} \cdot \text{m} \]Then, find the angular acceleration \( \alpha \):\[ \alpha = \frac{\tau}{I} = \frac{3 \times 10^{-4}}{4.5 \times 10^{-6}} = 66.67 \text{ rad/s}^2 \]

Determine Time to Decrease Rotational Speed

Using the equation for angular motion, \( \omega_f = \omega_i + \alpha t \), where \( \omega_f \) is the final angular velocity, \( \omega_i \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( t \) is the time. We know \( \omega_f = \omega_i - 22.5 \text{ rad/s} \).Rearrange this equation to solve for time:\[ t = \frac{\Delta \omega}{\alpha} = \frac{22.5}{66.67} = 0.337 \text{ s} \]

Final Check and Summary

Verify the calculations and ensure the units are consistent throughout the problem. The calculations give an angular acceleration \( \alpha = 66.67 \text{ rad/s}^2 \) and a time to decrease the speed by 22.5 rad/s of \( t = 0.337 \text{ s} \).

Key concepts

Moment of Inertia

The concept of moment of inertia is crucial in understanding how objects spin. Think of it as the rotational equivalent of mass in linear motion. It's a measure of how difficult it is to change the rotational speed of an object. For different shapes, the formulas for calculating moment of inertia vary.

In the exercise, we determine the moment of inertia for a solid sphere using the formula:

• \( I = \frac{2}{5}mr^2 \)

Here, \( m \) is the mass of the sphere, and \( r \) is the radius. This formula reflects the fact that mass distributed further from the axis makes an object harder to start or stop spinning.

For our problem, we've converted the mass to kilograms and radius to meters to use consistent units. By substituting these values into the formula, we find the sphere’s moment of inertia is \( 4.5 \times 10^{-6} \, \text{kg} \cdot \text{m}^2 \). Understanding and calculating moment of inertia helps in predicting how a given force will affect an object’s rotational motion.

Rotational Motion Equations

Rotational motion equations describe the behavior of rotating objects. They're similar to equations for linear motion but applied to angles and rotations.

One important equation used in this scenario is Newton's second law for rotation:

• \( \tau = I\alpha \)

Here, \( \tau \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. This formula shows how torque affects rotational acceleration, much like force affects linear acceleration.

Another key equation used is for calculating changes in angular speed over time:

• \( \omega_f = \omega_i + \alpha t \)

This helps determine how the angular velocity \( \omega \) changes with time \( t \). In this exercise, by rearranging the equation, we calculated the time it takes for the sphere’s rotational speed to decrease by a specific amount, given its angular acceleration.

These equations are critical tools in solving problems involving rotational dynamics.

Torque

Torque is the rotational force that causes an object to spin. It plays a role analogous to linear force in motion. Essentially, torque determines how much a force acting at a distance will rotate an object.

It is calculated as:

• \( \tau = fr \)

where \( f \) is the force acting perpendicular to the point of rotation, and \( r \) is the distance from the axis of rotation. This exercise involved calculating torque using the frictional force and the radius of the sphere.

The calculation\( \tau = 3 \times 10^{-4} \text{ N} \cdot \text{m} \) gives us the torque resulting from friction. With this torque, we showed how it affects the sphere's angular acceleration using Newton’s second law for rotation.

Torque is fundamental in understanding and controlling rotational movements in mechanical systems.