__Question 7.21__:(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:

KE

_{rot}= KE

_{trans}

(1/2) I ω

^{2}= (1/2)mv

^{2}

Energy of the cylinder at point B =

*m*g

*h*

Using the law of conservation of energy, we can write:

(1/2) I ω

^{2}= (1/2)mv

^{2}= mgh

Moment of inertia of the solid cylinder, I = (1/2)mr

^{2}

∴ 1/2 [ (1/2)mr

^{2}] + (1/2) mv

^{2}= mgh

(1/4)mr

^{2}ω

^{2}+ (1/2)mv

^{2}= mgh

But we have the relation, v = rω

∴ (1/4)v

^{2}+ (1/2)v

^{2}= gh

(3/4)v

^{2}= gh

∴ h = (3/4)v

^{2}/ g

= (3/4) X 5

^{2}/ 9.8 = 1.91 m

In ΔABC:

Sin θ = BC / AB

Sin 30

^{0}= 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 + (K

^{2}/ R

^{2}) ) ]

^{1/2}

∴ v = [ 2gABSinθ / (1 + (K

^{2}/ R

^{2}) ]

^{1/2}

For the solid cylinder, K

^{2}= R

^{2}/ 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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