A uniform solid cylinder of mass M = 0.50 kg and radius R = 0.10 m is released from rest, rolls without slipping: down a 1.0 m long inclined plane, and is launched horizontally from a horizontal table of height 0.75 m. The
inclined plane makes an angle of 30 degrees with the horizontal. The cylinder lands on the floor a distance D
away from the edge of the table, as shown in the figure. There is a smooth transition from the inclined plane to
the horizontal table, and the motion occurs with no frictional energy loss. The rotational inertia of a cylinder
around its center is MR2.
M -0.50 kg
R 0.10 m
1.0 m
30
Table
0.75 m
Floor
D
A. Calculate the total kinetic energy of the cylinder as it reaches the horizontal table
B. Calculate the angular velocity of the cylinder around its axis at the moment it reaches the floor.
C. Calculate the ratio of the rotational kinetic energy to the total kinetic energy for the cylinder at the moment it
reaches the floor.
D. Calculate the horizontal distance D
A sphere of the same mass the radius is now rolled down the same inclined plane. The rotational inertia of a
sphere around it center is 2/5 MR2.
E.
i. Is the total kinetic energy of the sphere at the moment it reaches the floor greater than, less than or equal to
the total kinetic energy of the cylinder at the moment it reaches the floor?
Greater than Less Than Equal to
Justify your