# Conservation of Energy Problem

1. Dec 2, 2006

### webdivx

1. The problem statement, all variables and given/known data
http://img172.imageshack.us/img172/5227/4rj8.jpg [Broken]

2. Relevant equations
Not sure, but I think KE=.5mv2 and PE=mgh are two possibly useful equations for this problem.

3. The attempt at a solution
I have no idea how to even start this problem. I've been searching the internet for hours and I seriously need help. Step by step instructions on how to do this are really really appriciated.

Last edited by a moderator: May 2, 2017
2. Dec 2, 2006

When using conservation of energy, you have to define the system which is of interest to you. You have a cylinder and a mass attached to the rope. Now, what exactly does conservation fo energy tell you, if you analyse the energies of the system at the point before the release and at the point just before the mass strikes the ground? What kind of energy do the objects in motion possess?

3. Dec 3, 2006

### webdivx

they possess Kinetic Energy if they are in motion.

4. Dec 3, 2006

Exactly. Note that the cylinder possesses rotational kinetic energy, since it has an angular velocity $$\omega$$. Find the relation between the translatoral velocity v of the other mass and between the angular velocity of the cylinder, and simply use energy conservation.

5. Dec 3, 2006

### webdivx

i don't know how to do that

6. Dec 3, 2006

Okay, do you know how to use conservation of energy? The total energy of the system is constant in time, so you can pick two points in time at which you will analyse the system. Let's pick the first point as the point just before the mass on the rope is left to move freely, and the second point as the point just before the same mass falls onto the ground. Now, define the energy of the system (i.e. the sum of potential and kinetic energy) for point 1 and for point 2, and then set them equal.

7. Dec 3, 2006

### webdivx

KE=.5mv2 and PE=mgh

point 1 = PE
point 2 = KE + PE

so it would be mgh = .5mv2 + mgh

8. Dec 3, 2006

There is no potential energy at point 2, since the mass is alomst on the floor. Further on, '.5mv2' (which is supposed to be 0.5mv^2, I assume), as you wrote, is the kinetic energy of the mass attached to the rope. What is the kinetic energy of the cylinder? You have to count that energy in, too.

9. Dec 4, 2006

### webdivx

is the kinetic energy of the cylinder .5mv^2 = .5MR^2

10. Dec 4, 2006

### Hootenanny

Staff Emeritus
Last edited by a moderator: Apr 22, 2017
11. Dec 5, 2006

### webdivx

i just don't understand. theres like a physics block in my mind. can somebody please solve for me.

Last edited by a moderator: Apr 22, 2017
12. Dec 5, 2006

### Hootenanny

Staff Emeritus
I'm afraid we won't solve it for you here; however, if you show some effort we will guide you through the problem. Now, the total energy of the system should be conserved, the pulley is frictionless so we can write the total energy of the pulley system as;

Total energy = Gravitation Potential Energy of mass + Kinetic Energy of the mass + rotational kinetic energy of the cylinder

Can you take the next step?