Problem 30 - University Physics with Modern Physics

Question

At \(t=3.00 \mathrm{s}\) a point on the rim of a \(0.200-\mathrm{m}\) -radius wheel has a tangential speed of 50.0 \(\mathrm{m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude 10.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00 \mathrm{s}\) and \(t=0\) . (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{s} ?\) (d) At what time will the radial acceleration equal \(g ?\)

Step-by-Step Solution

Answer

(a) 50 rad/s², (b) 150 rad/s at t=3s, 0 rad/s at t=0, (c) 225 rad, (d) t = 0.14 s.

Step 1: Understand the relationship between tangential and angular variables

Calculate angular quantities using given tangential quantities. Use the relationship between tangential speed \(v_t\) and angular speed \(\omega\) given by \(v_t = r \omega\), where \(r\) is the radius of the wheel.

Step 2: Calculate wheel's constant angular acceleration

The tangential acceleration \(a_t = 10.0 \ \mathrm{m/s^2}\) is related to the angular acceleration \(\alpha\) by \(a_t = r \alpha\). Hence, \(\alpha = \frac{a_t}{r} = \frac{10.0}{0.200} = 50.0\ \mathrm{rad/s^2}\).

Step 3: Calculate angular velocities at \(t=3.00\,\mathrm{s}\) and \(t=0\)

Initial angular speed \(\omega_0 = 0\) at \(t=0\). At \(t=3.00\,\mathrm{s}\), angular speed \(\omega = \omega_0 + \alpha t = 50.0 \times 3.00 = 150.0\,\mathrm{rad/s}\).

Step 4: Calculate the angular displacement between \(t=0\) and \(t=3.00\,\mathrm{s}\)

Using the formula \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\), where \(\omega_0 = 0\), \(\theta = \frac{1}{2} \times 50.0 \times (3.00)^2 = 225.0\,\mathrm{rad}\).

Step 5: Find time when radial acceleration equals gravitational acceleration

Radial acceleration \(a_r = \omega^2 r\). Set \(a_r = g = 9.8\,\mathrm{m/s^2}\). \(\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{0.200}} = 7.0 \ \mathrm{rad/s}\). \(\omega = \omega_0 + \alpha t\), solving gives \(t = \frac{7.0 + 0}{50.0} = 0.14\,\mathrm{s}\).

Key Concepts

Angular Velocity

Angular velocity refers to how fast an object rotates or revolves relative to a central point. It is often denoted by the symbol \( \omega \) and is typically measured in radians per second (rad/s). For example, if a point on the rim of a wheel moves with a tangential speed, we can calculate its angular velocity using the formula:

• \( v_t = r \omega \)

Here, \( v_t \) is the tangential speed, and \( r \) is the radius of the wheel. In the exercise, at \( t=3.00 \, \mathrm{s} \), the tangential speed is given as 50.0 \( \mathrm{m/s} \), and the wheel's radius is 0.200 m. By rearranging and solving for \( \omega \), we get:

• \( \omega = \frac{v_t}{r} = \frac{50.0}{0.200} = 250.0 \, \mathrm{rad/s} \)

Therefore, at \( t=3.00 \, \mathrm{s} \), the angular velocity of the wheel is 250.0 rad/s. Understanding this connection between tangential and angular variables is crucial for solving problems involving rotational motion.

Angular Acceleration

Angular acceleration describes the rate at which the angular velocity changes with time. It is represented by the symbol \( \alpha \) and usually measured in radians per second squared (rad/s\(^2\)). If a wheel is experiencing a change in its angular velocity, there is an angular acceleration present. It can be related to tangential acceleration as follows:

• \( a_t = r \alpha \)

In this exercise, we know the tangential acceleration \( a_t \) of the wheel is 10.0 \( \mathrm{m/s^2} \), and the radius \( r \) is 0.200 m. We can find \( \alpha \) through the formula:

• \( \alpha = \frac{a_t}{r} \)

By substituting the values, we find that:

• \( \alpha = \frac{10.0}{0.200} = 50.0 \, \mathrm{rad/s^2} \)

This means the wheel's angular acceleration is 50.0 rad/s\(^2\), signifying how quickly the wheel's rotation speed is decreasing.

Tangential Speed

Tangential speed, symbolized as \( v_t \), represents the linear speed of a point situated on a rotating object, specifically along the path's tangent. It is crucial to understand that although the entire wheel rotates, a point at the rim experiences linear movement due to this rotation. In terms of formula:

• \( v_t = r \omega \)

Where \( \omega \) is the angular speed and \( r \) is the radius. Tangential speed directly ties the linear aspect of rotational motion to its angular counterpart, providing a complete picture of the object's motion in space. If we take the given values from the problem, with \( r = 0.200 \, \mathrm{m} \) and \( \omega \) obtained previously:

• At \( t = 3.00 \, \text{seconds} \), \( v_t = 0.200 \, \times \, 250.0 = 50.0 \, \text{m/s} \)

Understanding the relationship between tangential speed and other angular parameters helps us explore concepts such as the linear velocity a point on a rotating body experiences.

Radial Acceleration

Radial acceleration, also known as centripetal acceleration, is the acceleration directed towards the center of the circular path of a rotating object. It is vital for maintaining an object's circular motion and is calculated using:

• \( a_r = \omega^2 r \)

Given that it depends on the square of the angular velocity, radial acceleration increases significantly with rotation speed. In this task, the radial acceleration equals the gravitational acceleration \( g \) when:

• \( \omega = \sqrt{\frac{g}{r}} \)

Substituting known values, \( g = 9.8 \, \mathrm{m/s^2}, \) and \( r = 0.200 \, \mathrm{m}, \) we find:

• \( \omega = \sqrt{\frac{9.8}{0.200}} = 7.0 \, \mathrm{rad/s} \)

At this angular velocity, the radial acceleration will be equal to the acceleration due to gravity. By understanding radial acceleration, students gain insights into forces acting on an object in circular motion.