Question

A mass of 3.2 kg is rotating on a circular path of radius 0.6 m with an angular velocity of 66 rad/s. The new radius of the path is 0.8 m. What is the new value of angular velocity?

• A. 66 rad/s
• B. 37.125 rad/s
• C. 13.75 rad/s
• D. 3.2 rad/s

Answer 1

The correct option is B

37.125 rad/s

$\text{Mass, m = 3.2 kg ; Radius of the circular path,}{r}_1=0.6mAngularvelocity,{\omega }_1=66rad/s;Let{I}_{1}bethemomentofintertia.Angularmomentum,L_1={\omega }_1{I}_1=m{r}_1^2{\omega }_1----------\left(1\right)Now,Radiusofthepath,r_2=0.8mLetamgularvelocitybe{\omega }_{2}andmomentofinertiabe{I}_2.\therefore Angularmomentum,L_2={\omega }_2{L}_2=m{r}_2^2{\omega }_1----------\left(2\right)Equating\left(1\right)and\left(2\right),\Rightarrow m{r}_1^2{\omega }_1=m{r}_2^2{\omega }_2\Rightarrow {\omega }_2=\frac{r_1^2{\omega }_1}{r_2^2}=\frac{(0.6{)}^2\times 66}{(0.8{)}^2}=37.125rad/s$