#2 Parts A-E, please 2. (25 poiets) A solid cylinder of mass 0.500 kg and radius

Question

#2 Parts A-E, please 2. (25 poiets) A solid cylinder of mass 0.500 kg and radius 0.100 m is placed at the top of an inclined plane 2.00 m high and 5.00 m long. It starts from rest and rolls without slipping to the bottom of the incline. a) What is the moment of inertia of the cylinder? b) Using energy considerations find the translational (linear) speed of the cylinder at the bottom of the incline. c) What is the angular speed of the cylinder at the bottom of the incline? d) What is the angular momentum of the cylinder at the bottom of the incline?

Explanation / Answer

a)The moment of inertia will be:

I = 1/2 MR^2

I = 0.5 x 0.5 x 0.1^2 = 0.0025 kg-m^2

Hence, I = 0.0025 kg-m^2

b)The PE at the top get converted to KE of motion but

KE = KE trans + KE rot

KE = 1/2 m v^2 + 1/2 I w^2

w = v/R ; I = 1/2 m r^2

KE = 1/2 m v^2 + 1/2 x 1/2 m r^2 v^2/r^2 = 1/2 m v^2 + 1/4 m v^2 = 3/4 m v^2

KE = PE

3/4 m v^2 = m g h

v = sqrt (4gh/3) = sqrt (4 x 9.81 x 2/3) = 5.11 m/s

Hence, v = 5.11 m/s

c)w = v/R = 5.11/0.1 = 51.1 rad/s

Hence, w = 51.1 rad/s

d)L = I w

L = 0.0025 x 51.1 = 0.128 J-s

Hence, L = 0.128 J-s

e)I' = 2I = 2 x 0.0025 = 0.005 kg-m^2

from conservation of angular momentum

Li = Lf

I w = I' w'

w' = (I/I')w = (0.0025/0.005)51.1 = 25.55 rad/s

Hence, w' = 25.55 rad/s

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