Question

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

Short answer

(a) \(-50.0 \; \text{rad/s}^2\), (b) \(250 \; \text{rad/s}, 400 \; \text{rad/s}\), (c) \(975 \; \text{rad}\), (d) \(7.86 \; \text{s}\)

Step-by-step solution

Identify Key Variables

Identify and write out the known quantities: - Radius of the wheel, \( r = 0.200 \; \text{m} \).- Tangential speed at \( t = 3.00 \; \text{s} \), \( v_t = 50.0 \; \text{m/s} \).- Tangential acceleration, \( a_t = -10.0 \; \text{m/s}^2 \). (Note that the acceleration is negative because the wheel is slowing down.)

Calculate Angular Acceleration \(\alpha\)

Use the relation between tangential acceleration and angular acceleration:\[ a_t = r \alpha \]Rearrange to solve for \( \alpha \):\[ \alpha = \frac{a_t}{r} = \frac{-10.0 \; \text{m/s}^2}{0.200 \; \text{m}} = -50.0 \; \text{rad/s}^2 \].

Calculate Angular Velocity at \(t = 3.00 \; \text{s}\)

Use the relationship between tangential speed and angular velocity:\[ v_t = r \omega \]Rearrange to find \( \omega \) at \( t = 3.00 \; \text{s} \):\[ \omega = \frac{v_t}{r} = \frac{50.0 \; \text{m/s}}{0.200 \; \text{m}} = 250 \; \text{rad/s} \].

Calculate Angular Velocity at \(t=0\)

Use the formula for angular velocity with constant angular acceleration:\[ \omega = \omega_0 + \alpha t \]We already know \( \omega = 250 \; \text{rad/s} \) at \( t = 3.00 \; \text{s} \), so:\[ 250 = \omega_0 + (-50)(3) \]Solve for \( \omega_0 \):\[ \omega_0 = 250 + 150 = 400 \; \text{rad/s} \].

Calculate the Angle Turned by the Wheel

Use the kinematic equation for angular motion:\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]Substitute \( \omega_0 = 400 \; \text{rad/s} \), \( \alpha = -50 \; \text{rad/s}^2 \), and \( t = 3.00 \; \text{s} \):\[ \theta = (400)(3) + \frac{1}{2}(-50)(3)^2 \]\[ \theta = 1200 - 225 = 975 \; \text{rad} \].

Calculate the Time When Radial Acceleration Equals \(g\)

Radial acceleration is given by \( a_r = r \omega^2 \) and it should equal \( g = 9.8 \; \text{m/s}^2 \):\[ r \omega^2 = g \]\[ 0.200 \omega^2 = 9.8 \]\[ \omega^2 = \frac{9.8}{0.200} = 49 \]\[ \omega = 7 \; \text{rad/s} \]Use \( \omega = \omega_0 + \alpha t \):\[ 7 = 400 + (-50)t \]Solve for \( t \):\[ t = \frac{400 - 7}{50} = 7.86 \; \text{s} \].

Key concepts

Angular Acceleration

Angular acceleration is a crucial concept in understanding how rotational motion changes over time. It represents the rate at which the angular velocity of an object changes. In simpler terms, it's how quickly something speeds up or slows down as it spins.

For a wheel or any rotating object, the tangential acceleration (\( a_t \)) and the radius (\( r \)) can be used to find angular acceleration (\( \alpha \)). The relationship is given by the formula:

• \( a_t = r \alpha \)

This formula shows that angular acceleration is directly related to tangential acceleration and inversely related to the radius. In our exercise, the negative angular acceleration (\( \alpha = -50.0 \, \text{rad/s}^2 \)) indicates the wheel is slowing down. This occurs when the tangential acceleration is opposite to the direction of the motion.

Angular Velocity

Angular velocity measures how fast an object rotates or spins around a central point. Think of it as the rotational speed of an object. It is expressed in radians per second (\( \text{rad/s} \)).

To calculate the angular velocity (\( \omega \)), you can use the relationship between tangential speed (\( v_t \)) and the radius (\( r \)):

• \( v_t = r \omega \)

By rearranging the formula, we calculate that at \( t = 3.00 \, \text{s} \), the angular velocity \( \omega = 250 \, \text{rad/s} \).

Angular velocity can change over time if there is angular acceleration. Using the equation:

• \( \omega = \omega_0 + \alpha t \)

you find that the initial angular velocity (\( \omega_0 \)) at \( t = 0 \) was \( 400 \, \text{rad/s} \). This initial value tells us how fast the wheel was spinning before it started slowing down.

Tangential Speed

Tangential speed refers to the speed at which a point on the edge of a rotating object is moving along a path. It's the linear speed at the outer rim of the wheel or circle.

Given the radius (\( r \)) of a wheel and angular velocity (\( \omega \)), tangential speed (\( v_t \)) can be determined easily using:

• \( v_t = r \omega \)

In our example, at \( t = 3.00 \, \text{s} \), \( v_t = 50.0 \, \text{m/s} \). This value shows how fast the edge of the wheel moves in linear terms.

However, note that tangential speed varies if the angular speed of the wheel changes. The relationship ensures that any change in angular velocity directly affects tangential speed. This means, if the wheel accelerates or decelerates, the speed at its edge will also increase or decrease respectively.

Kinematic Equations

Kinematic equations for rotational motion help us predict different parameters of rotational systems such as a wheel. These equations give insights into displacement, velocity, and acceleration over time.

For angular motion, the equivalent of the linear motion equation \( s = ut + \frac{1}{2} a t^2 \) for rotation is:

• \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)

Here, \( \theta \) is the angular displacement, \( \omega_0 \) is the initial angular velocity, and \( \alpha \) is the angular acceleration.

In our scenario, using this equation, the wheel turns \( 975 \, \text{rad} \) between \( t = 0 \) and \( t = 3.00 \, \text{s} \). This calculation shows how much the wheel rotates during a given time when it is slowing down.