Question

suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. otherwise, we keep the experiment the same. compared to the pucks in the video, this pair of pucks will rotate. view available hint(s). hint 1. how to approach the problem. this question asks you to relate the mass of a rotating object to the objects angular speed ω (rate of rotation). you can do that using angular momentum: l = i ω. in this equation, the moment of inertia is the “mass” term, equivalent to m in the equation for linear momentum. so, how does the pucks moment of inertia i relate to their mass m? is the relationship linear (i ∝ m)? or is it quadratic (i ∝ m²) or inverse square (i ∝ 1/m)? to see the answer, try writing the general equation that defines i. now, how will doubling the mass of the pucks affect i, l, and ω? one - fourth as fast. at the same rate. four times as fast. twice as fast. one - half as fast.

Explanation:

Step1: Recall angular - momentum formula

The angular - momentum formula is $L = I\omega$, where $L$ is angular momentum, $I$ is the moment of inertia, and $\omega$ is the angular speed. In the absence of external torques, angular momentum is conserved, i.e., $L$ is constant.

Step2: Relate moment of inertia to mass

For a point - like object (or in a simple case where the pucks can be approximated as point - like for the purpose of moment - of - inertia calculation), the moment of inertia $I = mr^{2}$ (assuming the pucks rotate about a fixed axis and $r$ is the distance from the axis of rotation). If the size of the pucks remains the same and only the mass $m$ is doubled, the new moment of inertia $I'$ is related to the original moment of inertia $I$ by $I'=2m\times r^{2}=2I$ (since the mass $m$ is doubled).

Step3: Analyze the effect on angular speed

Since $L$ is constant ($L = I\omega=I'\omega'$), and $I' = 2I$, we can write $I\omega=2I\omega'$. Solving for $\omega'$, we get $\omega'=\frac{\omega}{2}$.

Answer:

one - half as fast.