A hollow ball is rolling along a horizontal surface at 3.7 m/s when it encounters an upward incline. If it rolls without slipping up the incline, what maximum height will it reach?
2007-10-21 16:01:13 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
With the hollow ball, you have both the Kinetic Energy of speed and the Kinetic Energy of rotation.
KE= KEv + KEr
KEv= Kinetic Energy of velocity
KEr = KE of rotation.
KEr=1/2*I*w^2
I=2/3 *M*R^2 = Moment of Inertia for a Hollow Ball
w= V/R
KEr=1/2*(2/3*M*R^2)*(V^2/R^2)
=1/3*MV^2
therefor
KE = KEv+KEr
=1/2 MV^2 + 1/3 MV^2
=5/6 MV^2
KE=PE
5/6*M*V^2 = M*g*h
Therefore:
h = 5/6 * V^2 * 1/g
2007-10-21 17:51:41 · answer #1 · answered by Frst Grade Rocks! Ω 7 · 1⤊ 0⤋
KE=PE
0.5*m*v^2 = m*g*h
Therefore:
h = v^2/(2*g)
OMG I forgot about rotational energy, good work Remo.
2007-10-21 16:10:22 · answer #2 · answered by Damian M 3 · 1⤊ 1⤋