System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.21

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?
Solution:
A solid cylinder rolling up an inclination is shown in the following figure.

Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, θ = 30°
Height reached by the cylinder = h
(a) Energy of the cylinder at point A:
KErot = KEtrans
(1/2) I ω2 = (1/2)mv2
Energy of the cylinder at point B = mgh
Using the law of conservation of energy, we can write:
(1/2) I ω2 = (1/2)mv2 = mgh
Moment of inertia of the solid cylinder, I = (1/2)mr2
∴ 1/2 [ (1/2)mr2] + (1/2) mv2 = mgh
(1/4)mr2ω2 + (1/2)mv2 = mgh
But we have the relation, v = rω
∴ (1/4)v2 + (1/2)v2 = gh
(3/4)v2 = gh
∴ h = (3/4)v2 / g
= (3/4) X 52 / 9.8  =  1.91 m
In ΔABC:
Sin θ = BC / AB
Sin 300 = h / AB
AB = 1.91 / 0.5  =  3.82 m
Hence, the cylinder will travel 3.82 m up the inclined plane.

(b) For radius of gyration K, the velocity of the cylinder at the instance when it rolls back to the bottom is given by the relation:
v = [ 2gh / (1 + (K2 / R2) ) ]1/2
∴ v = [ 2gABSinθ / (1 + (K2 / R2) ]1/2
For the solid cylinder, K2 = R2 / 2
∴ v = [ 2gABSinθ / (1 + (1/2) ) ]1/2
= [ (4/3)gABSinθ ]1/2
The time taken to return to the bottom is:
t = AB / v
= AB / [ (4/3)gABSinθ ]1/2  =  (3AB / 4g Sinθ)1/2
= (11.46 / 19.6)1/2 = 0.764 s
Therefore, the total time taken by the cylinder to return to the bottom is (2 × 0.764) 1.53 s.

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