Question

Two circular discs of moment of inertia $0.4{\text{kg-m}}^2$ and $0.6{\text{kg-m}}^2$ are rotating about their own axis (vertical axis passing through geometric centre) at the rate of $200\text{rpm}$ and $50\text{rpm}$ both in clockwise direction. If both are gently brought in contact to rotate about same axis, the angular speed of the combination is

• A. $11.5\text{rad/sec}$
• B. $23\text{rad/sec}$
• C. $0\text{rad/sec}$
• D. $5.5\text{rad/sec}$

Answer 1

The correct option is A $11.5\text{rad/sec}$

Moment of inertia for the discs are:
$I_1=0.4{\text{kg m}}^2$
$I_2=0.6{\text{kg m}}^2$
Frequency of rotation for disc is:
$N_1=200\text{rpm},N_2=50\text{rpm}$
Hence angular speed of the discs about their axis of rotation:
${\omega }_1=\frac{2\pi N_1}{60}=\frac{20\pi }{3}\text{rad/s}$
${\omega }_2=\frac{2\pi N_2}{60}=\frac{5\pi }{3}\text{rad/s}$
Final moment of inertia $I=I_1+I_2$
$\Rightarrow I=1.0{\text{kg m}}^2$
Applying angular momentum conservation on system of discs, since torques of friction is internal and other external torque is absent.
$L_i=L_f$
$I_1{\omega }_1+I_2{\omega }_2=I{\omega }_f....\left(i\right)$
Putting the values in eq $\left(i\right)$:
$(0.4\times \frac{20\pi }{3})+(0.6\times \frac{5\pi }{3})=1\times {\omega }_f$
$\therefore {\omega }_f=\frac{11\pi }{3}=11.5\text{rad/s}$