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Question 7. 21. A solid cylinder rolls up an inclined plane of angle of inclination 30°. 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?


Velocity of the solid cylinder, v = 5 m/s

Angle of inclination, θ = 30°

Height reached by the cylinder =h



Energy of the cylinder at point A:

KErot=KE trans 12Iω2=12mv2
Energy of the cylinder at point B = mgh

Using the law of conservation of energy, 

12Iω2+12mv2=mgh
 12(12mr2)ω2+12mv2=mgh=14mr2ω2+12mv2=mgh14v2+12v2=gh        (as, v=rω)34v2=ghh=34v2g=34×5×59.8=1.91m
In ΔABC:

sinθ=BCABsin30=hABAB=1.910.5=3.82m

Hence, the cylinder will go 3.82 m up the inclined plane.

(b) For a radius of gyration K, the velocity of the cylinder at the instance when it rolls back to the bottom is,

v=(2gh1+K2R2)12 v=(2gABsinθ1+K2R2)12 For the solid cylinder,  K2=R22
 v=(2gABsinθ1+12)12=(43gABsinθ)12

The time taken to return to the bottom is,

t=ABv=AB(43gABsinθ)12=(3AB4gsinθ)12=(11.4619.6)12=0.764s

Therefore, the total time taken by the cylinder to return to the bottom=2×0.764 =1.53 s