Question

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

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

Answer 1

Answer: (B)

Given : Mass ($m$) = 2 kg
Initial radius of the path ($r_1$) = 0.8 m
Initial angular velocity ($\omega_1) = 44 \, rad\, s^{-1}$
Final radius of the path ($r_2$) = 1 m
Initial moment of inertia,
$I_{1 = mr_1^{2} = 2 \times\left(0.8\right)^{2} = 1.28\, kg \,m^{2}}$
Final moment of inertia $I_2 = mr_2^2 = 2 \times (1)^2$
$= 2 \, kg \, m^2$
From the law of conservation of angular momentum, we get
$I_{1}\omega_{1}=I_{2}\omega_{2} $
$\omega_{2}= \frac{I_{1}\times\omega_{1}}{I_{2}}=\frac{1.28\times44}{2} = 28.16 \, rad \, s^{-1}$