Question:
A solid cylinder rolls up an inclined plane of angle of inclination 30 degree . At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s. (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom?
Answer:
Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, θ = 30°
Let the cylinder go up the plane upto a height h.
Angle of inclination, θ = 30°
Let the cylinder go up the plane upto a height h.
From, 1/2 mv2 + 1/2 I ω2 = mgh
1/2mv2 + 1/2 (1/2 mr2) ω2 = mgh
3/4 mv2 = mgh
h = 3v2/4g = 3 × 52/4 × 9.8 = 1.913 m
If s is the distance up the inclined plane, then as
sin θ = h/s
s = h/sin θ = 1.913/sin 30° = 3.826 m
Time taken to return to the bottom
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