A circular disc of mass 300 gm and radius 20 cm can rotate freely about a vertical axis passing through is center O. A small insect of mass 100 gm is initially at a point A on the rim. The insect initially stationary starts walking from rest along the rim of disc with such a time varying relative velocity that the disc rotates in the opposite direction with a constant angular acceleration = 2πrad/s2. After some time T, the insect is back at the point A
. What is the time taken by insect to reach the original position?
Let the force applied by insect on the disc be F.
It is clear from the contact,
R×F=Idαd
R×F=MR22×2π
F=πMdR ..........(1)
The same amount of force will be applied to the insect by the disc,
Considering its torque about the centre of disc,
πMdR×R=Iiαii=insect
MiR2αi=MdπR2
αi=MdMi×π{MdMi=3}
αi=3π
now with respect to disc's frame of reference, αid=5π {Both α are opposite in direction}
now relative angular displacement = 2π
⇒2π=12αidt2⇒t=√54=√52s