A figure skater is spinning with an angular velocity of
. She then comes to a stop over a brief period of time. During this time, her angular displacement is +5.1 rad. Determine (a) her average angular acceleration and (b) the time during which she comes to rest.
The Correct Answer and Explanation is:
To solve this problem, we will assume the skater starts with an angular velocity ω0\omega_0ω0, comes to rest (final angular velocity ω=0\omega = 0ω=0), and experiences a constant angular acceleration while undergoing an angular displacement θ=+5.1 rad\theta = +5.1 \, \text{rad}θ=+5.1rad.
Since the original angular velocity is not given in the problem, let us denote it as ω0\omega_0ω0, and apply the kinematic equations for rotational motion:
(a) Average Angular Acceleration
We use the equation: ω2=ω02+2αθ\omega^2 = \omega_0^2 + 2\alpha\thetaω2=ω02+2αθ
Given that ω=0\omega = 0ω=0 (since the skater stops), we substitute: 0=ω02+2α(5.1)0 = \omega_0^2 + 2\alpha(5.1)0=ω02+2α(5.1)
Solve for α\alphaα: α=−ω022⋅5.1\alpha = -\frac{\omega_0^2}{2 \cdot 5.1}α=−2⋅5.1ω02
This is the expression for the average angular acceleration. It is negative because the skater is decelerating.
(b) Time to Come to Rest
We use the equation: ω=ω0+αt\omega = \omega_0 + \alpha tω=ω0+αt
Again, since ω=0\omega = 0ω=0: 0=ω0+αt⇒t=−ω0α0 = \omega_0 + \alpha t \Rightarrow t = -\frac{\omega_0}{\alpha}0=ω0+αt⇒t=−αω0
Substitute the expression for α\alphaα: t=−ω0−ω022⋅5.1=2⋅5.1ω0t = -\frac{\omega_0}{-\frac{\omega_0^2}{2 \cdot 5.1}} = \frac{2 \cdot 5.1}{\omega_0}t=−−2⋅5.1ω02ω0=ω02⋅5.1
So the time is: t=10.2ω0t = \frac{10.2}{\omega_0}t=ω010.2
Explanation
In rotational motion, angular displacement, angular velocity, and angular acceleration are related just as their linear counterparts are. In this case, the skater starts with an angular velocity ω0\omega_0ω0 and eventually comes to rest. The angular displacement she undergoes during this deceleration is 5.1 radians.
To find her average angular acceleration, we use the kinematic relation involving final angular velocity, initial angular velocity, angular acceleration, and angular displacement. Since she stops spinning, her final angular velocity is zero. Plugging this into the equation ω2=ω02+2αθ\omega^2 = \omega_0^2 + 2\alpha\thetaω2=ω02+2αθ, we solve for α\alphaα, the angular acceleration. The result is a negative value, indicating the skater is slowing down. The negative sign corresponds to a deceleration — a reduction in angular speed.
To find the time it takes her to stop, we use the relation between angular acceleration and time: ω=ω0+αt\omega = \omega_0 + \alpha tω=ω0+αt. Again, with a final angular velocity of zero, we solve for time ttt, which involves dividing the initial velocity by the magnitude of the angular acceleration. The resulting formula t=10.2ω0t = \frac{10.2}{\omega_0}t=ω010.2 shows that the time to stop is inversely proportional to how fast she was originally spinning. A higher initial spin means she would take less time to stop if the same displacement occurs.
