Question

A 2 kg mass is rotating on a circular path of radius 0.8 m with angular velocity of $44rad{s}^{-1}$ . If radius of path becomes 1 m, then what will be the value of angular velocity?

• A. $19.28rad{s}^{-1}$
• B. $28.16rad{s}^{-1}$
• C. $8.12rad{s}^{-1}$
• D. $35.26rad{s}^{-1}$

Answer 1

Answer: (B)

Given : Mass ($m$) = 2 kg
Initial radius of the path ($r_1$) = 0.8 m
Initial angular velocity (${\omega }_1)=44rad{s}^{-1}$
Final radius of the path ($r_2$) = 1 m
Initial moment of inertia,
$I_{1=m{r}_1^2=2\times {\left(0.8\right)}^2=1.28kg{m}^2}$
Final moment of inertia $I_2=m{r}_2^2=2\times (1{)}^2$
$=2kg{m}^2$
From the law of conservation of angular momentum, we get
$I_1{\omega }_1=I_2{\omega }_2$
${\omega }_2=\frac{I_1\times {\omega }_1}{I_2}=\frac{1.28\times 44}{2}=28.16rad{s}^{-1}$