Answer to Question #95750 in Mechanics | Relativity for Kshamaa.S

Question

A mass of 200 gm is attached to a cord wound over a flywheel of mass 10 kg and
circumference 75 cm with an axle diameter 3 cm. A mark is made at a distance 2 cm in
radial direction from the edge of the flywheel. Calculate the angular velocity and linear
velocity of this mark when the time is 20 s after the mass is released from the rest.

Answer 1

Flywheel angular acceleration $\epsilon = \frac{Rmg}{J+mR^2} = 0.51(s^{-2})$

which means the angular velocity is $\omega = \epsilon t = 0.51\cdot20 = 10.2 (s^{-1})$

and the linear velocity is $v = \omega (R - d) = 10.2\cdot 0.73 = 7.4 (m/s)$