Answer to Question #174067 in Mechanics | Relativity for Abdallah Muhammad abubakar

Question

A large disc has mass 2kg,radius 0.2m and initial angular velocity 50rad/s and a small disc of mass 4 kg,radius0.1m and initial angular velocity 200rad/s.find the common final angular velocity after the disc are pushed into contact:is the kinetic energy conserved during this process?

Answer 1

I will solve the question assuming that both the disk are rotating in same direction.

$m_1=2\;kg \\ r_1=0.2 \; m \\ I_1= \frac{1}{2}mr^2 = (0.2)^2= 0.04 \\ m_2=4 \; kg \\ r_2=0.1 \;m \\ I_2= \frac{1}{2}mr^2 = 2 \times (0.1)^2= 0.02$

Let the final angular velocity be w.

Since the external torque on system is 0. The angular momentum of the system will be conserved.

$I_1 \times w_1+I_2 \times w_2 = (I_1+I_2) \times w \\ 0.04 \times 50+0.02 \times 200=(0.04+0.02) \times w \\ 6=0.06 \times w \\ w=100 \; rad/sec \\ Energy \; initial = \frac{1}{2}I_1 \times w_1^2+\frac{1}{2}I_2 \times w_2^2= 50+400= 450 \\ Energy \; final =\frac{1}{2}I_1 \times w^2+\frac{1}{2}I_2 \times w^2= 300$

Kinetic Energy lost $= 450-300 = 150 \; J$ (energy is lost in the work done by kinetic friction)

No the kinetic energy will not be conserved.