Question

Diagram shows the top-view of an annular disc of mass $8\text{kg}$ that can freely rotate about its centre. With a cat of mass $2\text{kg}$ on its outer edge, it is initially rotating at an angular speed of $8\text{rad/s}$. What will be the increase in kinetic energy of the cat $\&$ ring system, if cat finally reaches to the inner edge.

• A. $25\text{J}$
• B. $80\text{J}$
• C. $10\text{J}$
• D. $39\text{J}$

Answer 1

The correct option is D $39\text{J}$

mass of annular disc, $M=8\text{kg}$,
Mass of cat, $m=2\text{kg}$
Outer radius, $R_2=0.8\text{m}$
Inner radius, $R_1=0.4\text{m}$
Initial angular speed, ${\omega }_i=8\text{rad/s}$
$\Rightarrow$When cat is at outer edge,the moment of inertia of (cat $\&$ disc) system
$I_i=m{R}_2^2+\frac{M(R_2^2+R_1^2)}{2}$
$I_i=2\times {0.8}^2+\frac{8({0.8}^2+{0.4}^2)}{2}$
$\therefore I_i=4.48{\text{kg.m}}^2$
Similarly, final moment of inertia when cat is at inner edge
$I_f=m{R}_1^2+\frac{M(R_2^2+R_1^2)}{2}$

$I_f=2\times {0.4}^2+\frac{8({0.8}^2+{0.4}^2)}{2}$
$\therefore I_f=3.52{\text{kg.m}}^2$
${\tau }_{\text{ext}}=0$,since forces acting between cat and annular disc is internal
hence, $L_i=L_f$
$\Rightarrow I_i{\omega }_i=I_f{\omega }_f$
$\Rightarrow 4.48\times 8=3.52\times {\omega }_f$
$\Rightarrow {\omega }_f=10.18\text{rad/s}$
Since, system only contain $K{E}_{\text{Rot}}$,
$\Rightarrow \Delta KE=K{E}_f-K{E}_i$
$\Delta KE=\frac{1}{2}{I}_f({\omega }_f{)}^2-\frac{1}{2}{I}_i({\omega }_i{)}^2$
$\Delta KE=\frac{1}{2}\times 3.52\times (10.18{)}^2-\frac{1}{2}\times 4.48\times (8{)}^2$
$\therefore \Delta KE\simeq 39\text{J}$
Here, $+ve$ sign represents increase in kinetic energy.