A solid cylinder and a hoop both of mass 0.2 kg and radius 1.0 cm are released f

Question

A solid cylinder and a hoop both of mass 0.2 kg and radius 1.0 cm are released from rest at the top of a 45 degree incline whose height is 1.0 m. Both roll without slipping to the bottom of the incline and continue to roll without slipping on a horizontal surface with the speed each reached at the bottom of the incline. The solid cylinder and the hoop then encounter another incline, this one making an angle 30 degree with the horizontal. Both move up the second incline without slipping. Calculate the center of mass speed of the solid cylinder and the hoop when they reach the bottom of the first incline. Find the distance along the second incline that the solid cylinder and the hoop will travel before they roll back down.

Explanation / Answer

a)

for solid cylinder :

h = height = 1 m

Using conservation of energy

Potential energy at Top = Kinetic energy + rotational Kinetic energy at Bottom

mgh = (0.5) m v2 + (0.5) I w2

mgh = (0.5) m v2 + (0.5) (0.5) (m r2) (v/r)2

mgh = (0.75) m v2

9.8 x 1 = (0.75) v2

v = 13.1 m/s

For Hoop :

h = height = 1 m

Using conservation of energy

Potential energy at Top = Kinetic energy + rotational Kinetic energy at Bottom

mgh = (0.5) m v2 + (0.5) I w2

mgh = (0.5) m v2 + (0.5) (m r2) (v/r)2

mgh = m v2

9.8 x 1 = v2

v = 9.8 m/s

b)

for solid cylinder :

h = height reached on 30 degree incline = 1m

d = distance travelled on the incline = h/Sin30 = 2 m

they both travel same length

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