Question 469: Centre of Mass and Rotational Motion

Question

A circular disc of mass M and radius R is rotating about its axis with angular speed $${\omega _1}$$
. If another
stationary disc having radius $${R \over 2}$$ and same mass M is droped co-axially on to the rotating disc.
Gradually both discs attain constant angular speed $${\omega _2}$$
 the energy lost in the process is p% of the
initial energy. Value of p is __________.

Accepted Answer

$${I_f}{\omega _f} = {I_i}{\omega _i}$$
$${I_i} = {{M{R^2}} \over 2}$$
$${I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}$$
$$ = {5 \over 4}.{{M{R^2}} \over 2}$$
$$\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} ight)}^2}} ight]\omega ' = \left( {{{M{R^2}} \over 2}} ight).\omega $$
$$ \Rightarrow $$ $$\left[ {{{M{R^2}} \over 2}.\left( {{5 \over 4}} ight)} ight]\omega ' = {{M{R^2}} \over 2}\omega $$
$$\omega  = {4 \over 5}\omega $$
loss of K.E. = $${{Loss} \over {{K_i}}} 	imes 100 $$
= $${{{1 \over 2}I{\omega ^2} - {1 \over 2}\left( {{5 \over 4}I} ight){{\left( {{4 \over 5}\omega } ight)}^2}} \over {{1 \over 2}I{\omega ^2}}}$$ $$ 	imes $$ 100
= $${{{\omega ^2} - {{16} \over {25}}{\omega ^2}\left( {{5 \over 4}} ight)} \over {{\omega ^2}}}$$ $$ 	imes $$ 100 = $$\left( {1 - {{80} \over {100}}} ight) 	imes 100$$
= 20%

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type