An ice skater has a moment of inertia of $5.0 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ when her arms are outstretched, and at this time she is spinning at 3.0 rev/s. If she pulls in her arms and decreases her moment of inertia to 2.0 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ how fast will she be spinning?
A) 7.5 rev/s
B) 3.3 rev/s
C) 10 rev/s
D) 2.0 rev/s

Question

An ice skater has a moment of inertia of $5.0 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ when her arms are outstretched, and at this time she is spinning at  3.0 rev/s. If she pulls in her arms and decreases her moment of inertia to  2.0  $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ how fast will she be spinning?
A)  7.5 rev/s
B)  3.3 rev/s
C)  10 rev/s
D)  2.0 rev/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 angular momentum $L$ is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. Initially, $L_i = I_i\omega_i = 5.0 \, \mathrm{kg\cdot m^2} \times 3.0 \, \mathrm{rev/s}$. After pulling her arms in, $I_f = 2.0 \, \mathrm{kg\cdot m^2}$ and we need to find the new angular velocity $\omega_f$. Setting the initial angular momentum equal to the final angular momentum gives $5.0 \times 3.0 = 2.0 \times \omega_f$, solving for $\omega_f$ gives $\omega_f = 7.5 \, \mathrm{rev/s}$.

An Ice Skater Has a Moment of Inertia Of 5.0 kg⋅m25.0 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }5.0 kg⋅m2

An Ice Skater Has a Moment of Inertia Of $5.0 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$