Question

A disc of radius $0.5m$ and mass $2kg$ rotates in the horizontal plane with its axis as axis of rotation at the rate of $600rev/min$. A ring of mass $1kg$ and internal radius $40cm$ and external radius $0.5m$ is kept on it with its axis coinciding with that of the disc. The speed of rotation now is: (in $rev/s$)

• A. $2.25$
• B. $10.5$
• C. $5.5$
• D. $1.265$

Answer 1

The correct option is C $5.5$

Moment of inertia of disc,

$I_{=}\frac{1}{2}m{r}^2=\frac{1}{2}2(0.5{)}^2=0.25kg{m}^2$

Moment of inertia of annular ring of inner radius $r_1$ and outer radius $r_2$,

$I_2=\frac{1}{2}{m}^{\prime }(r_1^2+r_2^2)=\frac{1}{2}\times 1\times ({0.4}^2+{0.5}^2)=0.205kg{m}^2$

By conservation of angular momentum,

$(I_1+I_2){\omega }_2=I_1{\omega }_1$

$\therefore {\omega }_2=\frac{I_1{\omega }_1}{I_1+I_2}$

Given,

${\omega }_1=600revspermin=10revspersec$

Substituting the values

${\omega }_2=\frac{0.25\times 10}{0.25+0.205}=\frac{2.5}{0.455}=5.5revpersec$