ke = .5 m v^2 + .5 I w^2
I = m r^2
w = v/r
so
ke = .5 m v^2 + .5 m r^2 v^2/r^2
ke = m v^2
v = 2.1 * 3.8 = 8.0 m/s
so
ke = 5.6 *64 Joules
= m g h = 5.6 g h
so
64 = g h
h = 64/9.8
sin 17.7 = h/distance up ramp
so
distance up ramp = h.sin 17.7
As part of a kinetic sculpture, a 5.6 kg hoop with a radius of 3.8 m rolls without slipping.
If the hoop is given an angular speed of 2.1 rad/s while rolling on the horizontal and then rolls up a ramp inclined at 17.7◦ with the horizontal, how far does the hoop roll along the incline? The acceleration of gravity is 9.81 m/s2 .
Answer in units of m
1 answer
1 answer