A figure skater is spinning with an angular velocity of +17.8 rad/s. She then comes to a stop over a brief period of time. During this time, her angular displacement is +5.14 rad. Determine (a) her average angular acceleration and (b) the time during which she comes to rest.
The Correct Answer and Explanation is:
We are given:
- Initial angular velocity: ωi=+17.8 rad/s\omega_i = +17.8 \, \text{rad/s}ωi=+17.8rad/s
- Final angular velocity: ωf=0 rad/s\omega_f = 0 \, \text{rad/s}ωf=0rad/s (since she comes to rest)
- Angular displacement: θ=+5.14 rad\theta = +5.14 \, \text{rad}θ=+5.14rad
We will use the rotational kinematics equation:ωf2=ωi2+2αθ\omega_f^2 = \omega_i^2 + 2\alpha\thetaωf2=ωi2+2αθ
(a) Finding Average Angular Acceleration α\alphaα
Solving for α\alphaα:0=(17.8)2+2α(5.14)0 = (17.8)^2 + 2\alpha(5.14)0=(17.8)2+2α(5.14)0=316.84+10.28α0 = 316.84 + 10.28\alpha0=316.84+10.28αα=−316.8410.28≈−30.84 rad/s2\alpha = \frac{-316.84}{10.28} \approx -30.84 \, \text{rad/s}^2α=10.28−316.84≈−30.84rad/s2
Answer (a): The average angular acceleration is −30.84 rad/s².
(b) Finding the Time to Come to Rest ttt
Use the formula:ωf=ωi+αt\omega_f = \omega_i + \alpha tωf=ωi+αt0=17.8+(−30.84)t0 = 17.8 + (-30.84)t0=17.8+(−30.84)tt=17.830.84≈0.577 secondst = \frac{17.8}{30.84} \approx 0.577 \, \text{seconds}t=30.8417.8≈0.577seconds
Answer (b): The time during which she comes to rest is 0.577 seconds.
Explanation
To solve this problem, we apply principles of rotational motion which mirror linear motion concepts. Just as linear velocity changes with acceleration over a certain distance, angular velocity changes with angular acceleration over an angular displacement. In this case, the figure skater is initially rotating with an angular velocity of 17.8 radians per second. She eventually comes to a stop, so her final angular velocity is 0 radians per second. During this slowing down process, she rotates through 5.14 radians.
We begin by using the angular version of the kinematic equation that relates initial and final angular velocity, angular acceleration, and angular displacement. Substituting the known values, we solve for the average angular acceleration. The negative value indicates that the skater is decelerating — her rotation is slowing down.
After finding the angular acceleration, we use another rotational motion equation to determine the time it takes for the skater to come to rest. Using the linear form ωf=ωi+αt\omega_f = \omega_i + \alpha tωf=ωi+αt, and substituting in the known values, we isolate ttt, the time, and solve.
This entire approach demonstrates how rotational motion can be analyzed similarly to linear motion by replacing position with angular displacement, velocity with angular velocity, and acceleration with angular acceleration.
