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प्रश्न
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?
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उत्तर १
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`
उत्तर २
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.
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