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?
At bottom of inclined plane centre of mass has speed 5 m s-1 let this be v.
We need to find how far the solid cylinder moves up on the plane.
By law of conservation of energy, energy is conserved.
Initial energy of body is the completely Kinetic (at bottom), On moving up the plane its kinetic energy gradually decreases. At top most point, Kinetic energy is zero and energy is completely gravitational potential energy.
The axis of rotation of cylinder passes through point B, we know that velocity of point O (axis of cylinder) is v. If ω is angular velocity of cylinder then,
v = R ω
(R is Distance between axis of rotation and axis of cylinder)
Initial KE,
KEi = KEr + KEt
Where
KEr is rotational kinetic energy
KEt is translational kinetic energy
KEt = 1/2 mv2
KEr = 1/2 I ω2
= 1/2 (mR2/2) ω2 = 1/4 mv2 (since v = R ω)
∴ Total kinetic energy, KEi = 1/4 mv2 + 1/2 mv2
=3mv2/4
Let h be the maximum height attained.
Then potential energy at that point,
PEf = mgh
By law of conservation of energy,
KEi = PEf
3mv2/4 = mgh
h = 3v2/4g
We know,
V=5 ms-1
g=9.8 ms-2
Thus,
h = 3(5)2/4× 9.8
= 1.91 m
Thus, maximum height the body would reach is 1.91 m from ground.
By law of conservation of energy,
3mv2/4 = mgh
v = (4gh/3)1/2
Let h be maximum height attained and θ be angle of inclination.
Then, Length travelled along plane to reach maxim um height,
L = h/sin(θ)
When v is the velocity, the times taken to reach maximum height,
T = L/v
= h/sin(θ) ÷ (4gh/3)1/2
= (3h/4g)1/2 ÷ sin(θ)
= (3 × 1.91/ 4 × 9.8)1/2 / sin (30)
= 0.765 s
the time taken for the body to move up and reach the bottom twice this time that is,
t = 2 × 0.765= 1.53 s