This is a classic problem that uses the principle of conservation of angular momentum.

Solution:

When Student A pulls his arms closer to his body, his moment of inertia decreases. According to the principle of conservation of angular momentum, if no external torques are acting on the system, the angular momentum will remain constant.

Mathematically, angular momentum $L$ is given by: $L = I \cdot \omega$ where:

• $I$ is the moment of inertia,
• $\omega$ is the angular velocity.

Since angular momentum is conserved, we have: $I_1 \omega_1 = I_2 \omega_2$ where:

• $I_1$ and $\omega_1$ are the initial moment of inertia and angular velocity (with arms extended),
• $I_2$ and $\omega_2$ are the final moment of inertia and angular velocity (with arms pulled in).

When Student A pulls his arms in:

• $I_2 < I_1$ (since the mass is closer to the axis of rotation),
• To keep $I_1 \omega_1 = I_2 \omega_2$ constant, $\omega_2$ must increase as $I_2$ decreases.

Conclusion:

When Student A pulls his arms in, his angular velocity $\omega$ increases. This is because the decrease in moment of inertia requires an increase in angular velocity to maintain the same angular momentum.

Would you like further details or have questions?

Related Questions:

1. What would happen if Student A extended his arms again after pulling them in?
2. How does the distribution of mass affect the moment of inertia in a rotating body?
3. If friction is present in the system, how would that affect the conservation of angular momentum?
4. How can we calculate the exact change in angular velocity if we know the initial and final moments of inertia?
5. What are some practical applications of the conservation of angular momentum in real-world physics?

Tip:

Remember, pulling mass closer to the axis of rotation decreases the moment of inertia, leading to an increase in angular velocity for a closed system.