Question

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?

Answer 1

Here, θ= 30°, v = 5 m/ s

Let the cylinder go up the plane up to a height h.
From 1/2 mv² +1/2IW² = mgh

`1/2 mv^2 + 1/2(1/2mr^2)omega^2 = mgh`

`3/4mv^2 =  mgh`

`h = (3v^2)/(4g) = (3xx5^2)/(4xx9.8) = 1.913 m`

if s is the distance up the inclined plane, then as

`sin theta = h/s, s =  h/(sin theta) = 1.913/sin 30^@  3.856 m`

Time taken to return to the bottom

`t = sqrt((2s(1+k^2/r^2))/(g sin theta)) = sqrt((2xx 3826(1+1/2))/(9.8 sin 30^@)) = 1.53s`

Answer 2

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 Iomega^2 = 1/2 mv^2`

Energy of the cylinder at point B = mgh

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

`1/2Iomega^2 = 1/2mv^2 = mgh`

Moment of inertia of the solid cylinder, `I = 1/2 mr^2`

`:.1/2(1/2 mr^2)omega^2 + 1/2 mv^2 = mgh`

`1/4 mr^2 omega^2 + 1/2 mv^2 = mgh`

But we have the relation `v = romega`

`:.1/4v^2 + 1/2v^2 = gh`

`3/4 v^2 =gh`

`:.h = 3/4 v^2/g`

`= 3/4 xx (5xx5)/(9.8) = 1.91 m`

In `triangleABC`

`sin theta = (BC)/(AB)`

`sin 30^@ = 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 = ((2gABsin theta)/(1+K^2/R^2))^(1/2)`

For the soild cylinder, `K^2 = R^2/2`

`:.v = ((2gABsin theta)/(1+1/2))^(1/2)`

`= (4/3gABsin theta)^(1/2)`

The time taken to return to the bottom is:

`t = (AB)/v`

`= (AB)/(4/3gABsintheta)^(1/2) =((3AB)/(4gsintheta))^"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.