A small cube (m=0.290 kg) is at a height of 261 cm up a frictionless track which has a loop of radius, R = 28.71 cm at the bottom. The cube starts from rest and slides freely down the ramp and around the loop. Find the speed of the block when it is at the top of the loop.

Question

A uniform solid cylinder (m=0.290 kg, of small radius) is at the top of a similar ramp, which has friction. The cylinder starts from rest and rolls down the ramp without sliding and goes around the loop. Find the speed of the cylinder at the top of the loop.

Answer 1

height of top of loop = .2871*2 = .5742 meters
potential energy at top of loop = m g h = m(9.81)(.5742) = 5.63 m Joules
(note I am not using m, does not matter elephant or mouse)
potential energy at start = m(9.81)(2.61)= 25.604 m Joules

so loss of potential energy = 25.604m - 5.63 m = 20 m Joules
that is not going to friction because there is none so
(1/2) m v^2 = 20 m
v^2 = 40
v = 6.32 m/s
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let w = omega = angular velocity
If it is rolling the loss of PE is still 20 m

now though that is (1/2) I w^2 + (1/2) m v^2

but v = w r if not slipping

I = (1/2) m r^2
so
ke = (1/4) m r^2 (v^2/r^2) + (1/2)m v^2

= m (3/4) v^2
so
20 m = m (3/4) v^2
80/3 = v^2
v = 5.16 m/s