Sliding down a Track
A small cube (m=0.730 kg) is at a height of 321 cm up a frictionless track which has a loop of radius, R = 41.73 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.

(in m/s)


A uniform solid cylinder (m=0.730 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.


(in m/s)

Q1: Why is it important that both the cube and the cylinder radius are small?
A1: To make things simple. If they were comparable to the dimensions of the track then we could not ignore the fact that the center of mass of cube (or cylinder) is not at the track's level, but a bit above it (or sometimes below it).

Q2: Why is it important that the track is frictionless for the cube and has friction for the cylinder?
A2: No friction for the cube means that the friction force does no work and does not influence the cube's speed. Friction for the cylinder means that the cylinder can roll down the track, and no sliding means that its speed v and angular velocity w are always related by v=w*r (r - cylinder radius).

Q3: Would the answers for the cube and cylinder be different?
A3: Yes, because the same potential energy change for the cylinder is turned not only into the kinetic energy of motion of center of mass, but also into the kinetic energy of rotation (remember, v=w*r): Km + Kr. For the cube, the potential energy change is turned only into the kinetic energy of motion: Km.

Q4: What is the kinetic energy of rotation?
A4: Kr = (1/2)*I*w^2, here I is the cylinder's moment of inertia (look up definition): I (for uniform solid cylinder) = (1/2)*m*r^2 (r - radius of cylinder).

Q5: Should I use the energy conservation law?
A5: Yes, nothing else would be useful here. The potential energy change is turned into kinetic energy:

cube: m*g*(h-2R) = (1/2)*m*v^2;
cylinder: m*g*(h-2R) = (1/2)*m*v^2 + (1/2)*((1/2)*m*r^2)*(v/r)^2;

Looks like there is no need for the value of mass.

It seems to me that there is information missing in the question...

It is asking for the speed of the cube/cylinder at the top of the loop? :confused:

If they were released from the top of the loop, the answer would be zero, if not, then they would have some value of speed/kinetic energy at the bottom of the loop and the initial speed at the top of the loop will depend on that speed/kinetic energy at the bottom of the loop.

And when you said 'is at a height of 321 cm up a loop' does that mean that the loop is 321 cm high? Or is that loop higher, and at the moment the question was asked, the cube had already slid down some distance down the loop?

you really gave me the way to understand the question, I solved it without looking at the last part ; ) haha
thanks

I have a question
about this Kr = (1/2)*I*w^2, here I = (1/2)*m*r^2.. is it always right ? For all problems?

Yes, you are correct.