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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?
Asked by Mayur Last Modified
Answer 
Disha J K
Experienced Physics and Mathematics tutor
At the bottom of the inclined plane, the energy of the cylinder is kinetic,
Initial energy =
The angular velocity of an object rolling without slipping is related to its centre of mass speed by,
where r is the radius of the cylinder. The moment of inertia about the axis of rotation for a solid cylinder is,
. Substituting this in the above equation gives,
initial energy =
The cylinder rolles up and halts before it rolls down, the kinetic energy at that instant is zero. The enrgy is purely potential. If it has rolled a distance x up the plane, the height at that point is
The final energy =
Applying conservation of energy, the final energy = initial energy
(a) The cylinder will go 3.75m up the plane.
Assume that the plane is frictionless. Due to conservation of energy, the final centre of mass of cylinder would then be 5 m/s.
Initial velocity,
Then,
(b) It takes 1.5s to return to the bottom
read less
1)energy conservation.
rotational kinetic energy+linear kinetic energy=PE(at the top of the plane)
0.5*I*w^2+0.5*m*v^2=m*g*h
h=elevation on plane
distance travelled=h/sin30
distance travelled=3.8226
b)while on return, potential energy converted to Kinetic Energy.
velocity at bottom required to be calculated and use newtons law of motion
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