# Block on a roller coaster

## Homework Statement

A small cube (m=0.450 kg) is at a height of 393 cm up a frictionless track which has a loop of radius, R = 58.95 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.

v^2=g (0.66H-2R)

## The Attempt at a Solution

This was one of those questions that i never understood. The final is coming up and its bothering me that i still don't no how to do this.'
obviously the equation i tried above's not right... any help?

thanks!

its a conservation of energy problem, with only KE and PE
you know the total mechanical energy (Emec) and Emec=KE+PE=.5mv^2+mgh

oh crap posted wrong part >.<
The block's suppose to be "A uniform solid cylinder (m=0.450 kg, of small radius)" and answer's 6.00m/s. Sorry.

ok, that doesnt change the question too much.
you use the same equation as before, Emec=KE+PE, but now KE becomes the kinetic energy of a rolling object.
As you know the kinetic energy of a object moving linearly (like a block) is .5mv^2.
the kinetic energy of a rotating object is similar, .5Iw^2 (the w is supposed to be omega)

Therefore, the kinetic energy of a rolling object, one that rotates and moves is:
KE=.5mv^2+.5Iw^2

The problem is the same as before, except for the KE and the fact you know have to find the equations for moment of inertia of a solid cylinder and relate the angular velocity to linear velocity. Now, even though they don't give you the radius, if you set everything up properly, that shouldn't be a problem

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I got it! i didn't know that it was still a conservation problem.

thanks alot for the help