A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of $2.25 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ If he pulls in his arms so his moment of inertia decreases to $1.80 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ angular speed? what will be his new
A) 4.60 rad/s
B) 0.81 rad/s
C) 2.25 rad/s
D) 6.25 rad/s

Question

A figure skater rotating at  5.00 rad/s with arms extended has a moment of inertia of $2.25 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ If he pulls in his arms so his moment of inertia decreases to $1.80 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ angular speed? what will be his new
A)  4.60 rad/s
B)  0.81 rad/s
C)  2.25 rad/s
D)  6.25 rad/s

Quizplus · Accepted Answer

The conservation of angular momentum principle states that the initial angular momentum is equal to the final angular momentum when no external torque acts on the system. The formula for angular momentum is $L = I\omega$, where $L$ is the angular momentum, $I$ is the moment of inertia, and $\omega$ is the angular velocity. Initially, the skater has a moment of inertia $I_1 = 2.25 \, \mathrm{kg \cdot m^2}$ and an angular velocity $\omega_1 = 5.00 \, \mathrm{rad/s}$. After pulling his arms in, his moment of inertia decreases to $I_2 = 1.80 \, \mathrm{kg \cdot m^2}$. To find the new angular velocity $\omega_2$, we use the conservation of angular momentum: $I_1\omega_1 = I_2\omega_2$. Solving for $\omega_2$ gives $\omega_2 = \frac{I_1\omega_1}{I_2} = \frac{2.25 \cdot 5.00}{1.80} = 6.25 \, \mathrm{rad/s}$.

A Figure Skater Rotating at 5 2.25 kg⋅m22.25 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }2.25 kg⋅m2

A Figure Skater Rotating at 5 $2.25 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$