Question

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the centre of the disk. The platform has a mass of $150kg$, a radius of $2.0m$, and a rotational inertia of $300kg{m}^2$ about the axis of rotation. A $60kg$ student walks slowly from the rim of the platform toward the centre. If the angular speed of the system is $1.5rad/s$ when the student starts at the rim, what is the angular speed when she is $0.50m$ from the centre?

• A. 1.2 rad/s
• B. 2.6 rad/s
• C. 1.5 rad/s
• D. 3.6 rad/s

Answer 1

Answer: (B)

The initial rotational inertia of the system is $I_i=I_{\text{disk }}$ $+I_{\text{student }}$, where $I_{\text{disk }}=300kg{m}^2$ (which, incidentally, does agree with Table $10-2(c))$ and $I_{\text{student }}=m{R}^2$ where $m=60kg$ and $R=2.0m$.
The rotational inertia when the student reaches $r=0.5m$ is $I_f=I_{disk}+m{r}^2$. Angular momentum conservation leads to
$I_i{\omega }_i=I_f{\omega }_f\Rightarrow {\omega }_f={\omega }_i\left(\frac{I_{\text{disk }}+m{R}^2}{I_{\text{disk }}+m{r}^2}\right)$
which yields, for ${\omega }_i=1.5rad/s$, a final angular velocity of ${\omega }_f=2.6rad/s$