Problem 11 - University Physics with Modern Physics

Question

A machine part has the shape of a solid uniform sphere of mass 225 \(\mathrm{g}\) and diameter 3.00 \(\mathrm{cm} .\) 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 \(\mathrm{N}\) at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by 22.5 \(\mathrm{rad} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) Angular acceleration is 22.22 rad/s². (b) Time is 1.01 seconds.

Step 1: Convert mass and diameter to standard units

The mass of the sphere is given as 225 g, which needs to be converted to kilograms. Since 1 kg = 1000 g, we have:\[ \text{Mass} (m) = \frac{225}{1000} \text{ kg} = 0.225 \text{ kg}. \]The diameter is given as 3.00 cm. To find the radius (r) in meters, divide by 2 and convert centimeters to meters:\[ r = \frac{3.00}{2 \times 100} \text{ m} = 0.015 \text{ m}. \]

Step 2: Find the moment of inertia of the sphere

The formula for the moment of inertia (I) of a solid sphere about its center is:\[ I = \frac{2}{5} m r^2. \]Substitute the mass and radius found in Step 1 into this formula:\[ I = \frac{2}{5} \times 0.225 \text{ kg} \times (0.015 \text{ m})^2 = 1.35 \times 10^{-5} \text{ kg} \cdot \text{m}^2. \]

Step 3: Calculate the torque due to friction

Torque (τ) is given by the formula:\[ \tau = F \times r, \]where F is the friction force and r is the radius. Here, F = 0.0200 N and r = 0.015 m:\[ \tau = 0.0200 \text{ N} \times 0.015 \text{ m} = 3.00 \times 10^{-4} \text{ N} \cdot \text{m}. \]

Step 4: Determine the angular acceleration

Angular acceleration (α) is given by the formula:\[ \alpha = \frac{\tau}{I}, \]where τ is the torque and I is the moment of inertia. Substitute the values from Steps 2 and 3:\[ \alpha = \frac{3.00 \times 10^{-4} \text{ N} \cdot \text{m}}{1.35 \times 10^{-5} \text{ kg} \cdot \text{m}^2} \approx 22.22 \text{ rad/s}^2. \]

Step 5: Calculate the time to change rotational speed

Use the kinematic equation for angular motion:\[ \Delta \omega = \alpha \times t, \]where \( \Delta \omega = 22.5 \text{ rad/s} \) and \( \alpha \approx 22.22 \text{ rad/s}^2. \)Solve for time (t):\[ t = \frac{\Delta \omega}{\alpha} = \frac{22.5 \text{ rad/s}}{22.22 \text{ rad/s}^2} \approx 1.01 \text{ s}. \]

Key Concepts

Angular Acceleration

Angular acceleration (\( \alpha \)) describes how quickly an object's rotational speed changes. It's essentially the rotational equivalent of linear acceleration. In the exercise you're looking at, the machine part is subject to friction as it spins, which affects its speed.

In our context, angular acceleration tells us how fast the speed of the sphere spinning around its axle is decreasing due to the friction. To calculate this, we apply the formula:

• \( \alpha = \frac{\tau}{I} \)

Here, \( \tau \) represents the torque applied to the object and \( I \) is the object's moment of inertia.

The negative value of angular acceleration here would indicate a deceleration, because friction is slowing down the rotation. Understanding this decrease is crucial for predicting how long it takes for the sphere's rotation to diminish to a certain speed.

Moment of Inertia

The moment of inertia (\( I \)) is a property of a body that defines its resistance to angular acceleration. Think of it as the rotational equivalent of mass in linear dynamics. It tells you how much torque is needed to achieve a certain angular acceleration.

For this solid sphere, the formula used is:

• \( I = \frac{2}{5}m r^2 \)

This depends on both the mass of the sphere (\( m \)) and its radius (\( r \)). Knowing the moment of inertia helps calculate how the sphere will react to the frictional force trying to slow it down. A larger moment means more resistance.

This concept helps us determine the amount of torque necessary to change the spinning speed of the sphere, considering its mass distribution relative to its axis of rotation.

Torque

Torque (\( \tau \)) is the force that causes an object to rotate. For linear motion, we have force, and similarly, for rotational motion, we have torque. It essentially causes a turn or twist about an axis.

In the exercise, torque is produced by the frictional force acting at the equator of the spinning sphere. It's calculated using the formula:

• \( \tau = F \times r \)

where \( F \) is the force of friction and \( r \) is the radius of the sphere.

In simpler terms, the further the force is applied from the axis, the larger its effect (or leverage) and the greater the torque. Understanding torque is vital in finding how this acting friction impacts the sphere's rotation speed, leading to angular deceleration.

Friction Force

Friction force is the force that resists the motion of one surface against another. It's the 'drag' that prevents surfaces from sliding past each other without effort. Even when it is a small force, as in our exercise, it has a significant role.

Here, the sphere's edge rubs against metal, creating a friction force of 0.0200 N. This friction is the cause behind the torque that decelerates the sphere, thereby inducing angular acceleration in the opposite direction of its initial spin.

• Friction opposes motion; it converts kinetic energy into heat.
• It's responsible for slowing down moving parts, which is crucial in real-world applications.

By grasping the basics of friction force, you can better understand how it affects rotational dynamics beyond the surface level influence on wheels and spheres.