Question 591
CBSE Class 11 Physics · Systems of Particles and Rotational Motion · Rotational dynamics
A child on a frictionless turntable rotates at 40 rev min⁻¹ and reduces moment of inertia to \(\frac{2}{5}\) of its initial value. Find
(a)
the new angular speed
(b)
the kinetic-energy ratio.
Solution to Question 591
Complete workingWith zero external torque, angular momentum is conserved.
\[I_i\omega_i=I_f\omega_f\]
\[\begin{aligned}\omega_f &= \frac{I_i}{(\frac{2}{5})I_i}(40)\\&= 100\ \mathrm{rev\,min^{-1}}\end{aligned}\]
\[\begin{aligned}\frac{K_f}{K_i} &= \frac{I_f\omega_f^2}{I_i\omega_i^2}\\&= \frac{I_i}{I_f}\\&= \frac52\\&= 2.50\end{aligned}\]
Final answers by part
(a)
\[100\ \mathrm{rev\,min^{-1}}
(b)
\quad K_f=2.50K_i\]