A thin metal ring of diameter 0.6m and mass 1 kg starts from rest and rolls down an
inclined plane. Its linear velocity on reaching the foot of the plane is 5m/sec. Calculate (i)
moment of inertia of the ring and (ii) the kinetic energy of rotation at that instant,
Answers
Answer:
(i) M.O.I= 0.09 kg-m^2
(ii) Rotational K.E= 0.12505 Joule
Explanation:
(i) Formula for Moment of inertia of ring is (M) x(R^2). This is the M.O.I about a plane perpendicular to plane of ring.
Therefore, M.O.I = (1 kg) x ( 0.3^2)
= 0.09 kg-m^2
(ii) There will be two type of kinetic energy associated with a ring rolling down an incline plane. 1st is translational kinetic energy and 2nd is rotational kinetic energy. Since in the question we need to find only rotational K.E.
Therefore, Rotational K.E = (1/2)x(I)x(omega^2)
where,
I = moment of inertia of ring about a plane perpendicular to plane of ring.
omega = angular velocity of ring.
Formula that relates omega and linear velocity is,
V= (R)x(omega)
therefore,
omega = (5)/(0.3)
= 1.667 rad/s
Hence,
K.E = (1/2)x(0.09)x(1.667^2)
= 0.12505 Joule