Answer:
21.67 rad/s²
208.36538 N
Explanation:
= Final angular velocity = 
= Initial angular velocity = 78 rad/s
= Angular acceleration
= Angle of rotation
t = Time taken
r = Radius = 0.13
I = Moment of inertia = 1.25 kgm²
From equation of rotational motion
The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²
Torque is given by
Frictional force is given by
The magnitude of the force of friction applied by the brake shoe is 208.36538 N
It is gaining speed at the rate of 2 m/s per second.
I believe this question ask for the energy dissipated by
friction.
The overall energy equation for this is:
F = PE – KE
where F is friction loss, PE is potential energy = m g h,
KE is kinetic energy = 0.5 m v^2
<span>F = 66 kg * 9.8 m/s^2 * 170 m – 0.5 * 66 kg * (11 m/s)^2</span>
<span>F = 105,963 J ~ 106,000 J </span>
