Question

A hoop and a disk, both of 0.50- m radius and 4.0- kg mass, are released from the top of an inclined plane 2.9 m high and 8.7 m long. What is the speed of each when it reaches the bottom? Assume that they both roll without slipping. What is the speed of the hoop? What is the speed of the disk?

Answer 1

a) By using the conservation of energy

From Potential to Rotational and Translational KE:

For the hoop, PE = KE_r + KE_t

The rotational KE = 0.5 IW²

For the hoop, I = mr²

The relation between the linear velocity and angular velocity is V = r (or) = v / r

KE = 0.5*m r² V² / r²

= 0.5 mv²

Hence, mgh = 0.5 mv² + 0.5 mv²

=> gh = v²

v² = 9.81 * 2.9

v = 5.33 m/s

The speed of a hoop = 5.33 m/s

b)

The moment of inertia of the disk, I = 0.5 mr²

KE_r = 0.5* (0.5m r²) (v² / r²)

   = 0.25 m v²

mgh = 0.5 mv² + 0.5 mv²

9.8 * 2.9 = 0.75 v²

v² = 37.89

v = 6.15 m/s

The speed of a disk = 6.15 m/s