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?
Given: The angle of inclination of the plane is 30 ° and speed of the centre of the mass of the cylinder is 5 m/s .
a)
The moment of inertia of solid cylinder is given as,
I = 1 2 m r 2
Where, m is the mass of the solid cylinder and r is the radius of the solid cylinder.
The velocity of the centre of mass of the solid cylinder is given as,
v cm = r ω 0
Where, ω 0 is the angular velocity of the solid cylinder.
The initial kinetic energy of the solid cylinder is given as,
K 0 = 1 2 m v 0 2 + 1 2 I ω 0 2
By substituting the values of I and ω 0 in the above equation, we get
K 0 = 1 2 m v 0 2 + 1 2 ( 1 2 m r 2 ) ω 0 2 = 1 2 m v 0 2 + 1 4 m ( r ω 0 ) 2 = 1 2 m v 0 2 + 1 4 m ( v 0 ) 2 = 3 4 m v 0 2
Initial potential energy of the solid cylinder is zero as it is at the bottom of the inclined plane. Therefore,
U 0 = 0
In addition, at the final position on the inclined plane, the velocity of the cylinder becomes zero. Therefore, the final kinetic energy of the solid cylinder is zero.
K f = 0
The final potential energy of the solid cylinder is given as,
U f = m g h
Where, g is the gravitational acceleration and h is the vertical distance covered by the solid cylinder.
From the law of conservation of energy, we get
U 0 + K 0 = U f + K f 0 + 3 4 m v 0 2 = m g h + 0 3 4 m v 0 2 = m g h h = 3 v 0 2 4 g
By substituting the values in the above equation, we get
h = 3 × 5 2 4 × 9.81 = 1.91 m
Consider the figure shown below.
From the above figure, the distance covered by the solid cylinder along the inclined plane is,
s = h sin 30 ° = 1.91 0.5 = 3.82 m
Thus, the distance traveled by the cylinder up the inclined plane is 3.82 m .
b)
The time taken to complete the motion along inclined plane is given as,
t = 2 s ( 1 + k 2 r 2 ) g sin θ
Where, k is the radius of gyration and θ is the inclination angle.
For solid cylinder,
k 2 r 2 = 1 2
By substituting the values in the above equation, we get
t = 2 × 3.82 × ( 1 + 1 2 ) 9.81 × sin 30 ° = 1.53 sec
Same amount of time would require to reach the bottom of the inclined plane. Therefore, total time taken to return to the bottom is,
t T = 1.53 sec + 1.53 sec = 3.06 sec ≈ 3.0 sec
Thus, the time taken by cylinder to return to the bottom is 3.0 sec.
Given: The angle of inclination of the plane is
a)
The moment of inertia of solid cylinder is given as,
Where,
The velocity of the centre of mass of the solid cylinder is given as,
Where,
The initial kinetic energy of the solid cylinder is given as,
By substituting the values of
Initial potential energy of the solid cylinder is zero as it is at the bottom of the inclined plane. Therefore,
In addition, at the final position on the inclined plane, the velocity of the cylinder becomes zero. Therefore, the final kinetic energy of the solid cylinder is zero.
The final potential energy of the solid cylinder is given as,
Where,
From the law of conservation of energy, we get
By substituting the values in the above equation, we get
Consider the figure shown below.
From the above figure, the distance covered by the solid cylinder along the inclined plane is,
Thus, the distance traveled by the cylinder up the inclined plane is
b)
The time taken to complete the motion along inclined plane is given as,
Where,
For solid cylinder,
By substituting the values in the above equation, we get
Same amount of time would require to reach the bottom of the inclined plane. Therefore, total time taken to return to the bottom is,
Thus, the time taken by cylinder to return to the bottom is 3.0 sec.
