A small cube (m=0.530 kg) is at a height of 387 cm up a frictionless track which has a loop?

Question

of radius, R = 58.05 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.

A uniform solid cylinder (m=0.530 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

First one's pretty easy. Since you have a frictionless track (and we're assuming there is no air resistance here, either), the speed is due solely to the change in potential and kinetic energies. At the start, the cube has a potential energy of mgh, or 0.530 kgX0.387mX9,8 m/s^2 = 2.01 J. By the time it reaches the bottom of the loop, it has converted all of that potential energy into kinetic energy (since we're calling the bottom of the track to be zero altitude). As it travels up the loop, it gains potential energy again as it fights against gravity, slowing it down. At the top of the track, therefore, it has a potential energy of 0.30 J, meaning the kinetic energy must be the difference, or 1.71 J. Solving for velocity (KE = 1/2 mv^2), you get 2.54 m/s.

The second case is a little more tricky, in that it involves rolling friction. However, friction only plays a role in starting and stopping motion -- if the cylinder achieves pure rolling motion, the frictional component goes to zero because there will be no relative motion between the points of contact on the ramp and the cylinder (no sliding). If we assume that there is never any slipping along its path, then it should behave just like the cube sliding along a frictionless track, and the speed will still be 2.54 m/s. I could be wrong, but I think that's the answer.

Answer 2

when you consider that loop is on the bottom, conservation of power PE at excellent = KE at bottom of loop 0.seventy 3 * 9.80 one * 0.321 = 0.5 * 0.seventy 3 * v^2 v = 2.fifty one m/s ---answer area 2) when you consider that coefficient of friction replaced into not given, so answer.