Question

A body starts rotating from rest by the application of a torque of 40 Nm. The kinetic energy of the body thereby

increases to 1600 J in 2 seconds. Find the M.I. of the body.​

Answer 1

Answer:

T = 20 Nm

n = 60 revolutions in 60s = 1 r.p.s

t = 1 min = 60 seconds

Moment of inertia ( I ) = ?

Angular displacement thita = wt = 2π nt

= 2π × 1 × 60

= 120π rad

=thita = w•t + 1/2 at square

120π = 0 + 1/2 a × 60 square

a = 240π/ 60 square

= π/ 15 rad /s square

t = Ia

I = t/ a

= 20/π/15

= 95.48 kg m square

moment of inertia of the body is 95.48 kg m square.