Question
A solid sphere of weight 36.0 N rolls up an incline at an angle of 30.0o. At the bottom of
the incline the center of mass of the sphere has a translational speed of 4.90 m s-1.
Isphere = 2/5mr^2.
What is the kinetic energy of the sphere at the bottom of the incline?
How far does the sphere travel up along the incline?
the incline the center of mass of the sphere has a translational speed of 4.90 m s-1.
Isphere = 2/5mr^2.
What is the kinetic energy of the sphere at the bottom of the incline?
How far does the sphere travel up along the incline?
Answers
drwls
It rolls up the hill until the initial kinetic energy, which is both translational and rotational, equals the potential energy gained.
Call the initial speed V and the maximum height H.
Initial KE = (1/2)MV^2 + (1/2)Iw^2
and, since I = (2/5)MR^2 and w = V/R,
Initial KE = (1/2)MV^2 + (1/5)MV^2 = ?
Set that sum equal to M g H and solve for H. The distance it travels along the incline is H/sin 30
Call the initial speed V and the maximum height H.
Initial KE = (1/2)MV^2 + (1/2)Iw^2
and, since I = (2/5)MR^2 and w = V/R,
Initial KE = (1/2)MV^2 + (1/5)MV^2 = ?
Set that sum equal to M g H and solve for H. The distance it travels along the incline is H/sin 30
Vansf2700
Good