# How do I find rotational energy for a pulley system?

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1. May 30, 2015

### Preptopro

1. The problem statement, all variables and given/known data
So I have a horizontal pulley positioned at the edge of table with a mass of .2kg hanging down from a height of .76meters, the other end of the string is attached to a wooden block of mass .25kg that when the .2kg weight is dropped the wooden block is pulled towards the pulley .8 meters at a velocity of .92m/s. Now I need to find the GPE, translational, and rotational energy of the system. The objective is to have all 3 types of energies in the system.

2. Relevant equations
GPE = mgh

Translational energy= .5mv^2

Rotational energy = .5Iw^2 ?

3. The attempt at a solution
GPE would just be (.2kg)(9.8m/s^2)(.76m) = 1.5 Joules right?

Translational energy = .5(.25kg)(.92m/s)= .11 Joules, this doesn't seem right I'm guessing I have to account for friction or something else? Any ideas?

And for Rotational energy I have no idea where to begin, I is .5MR^2 and then since w is (v/r) does that mean I could use .5(.5MR^2)(w^2) which is .25(MR^2)(v/r)^2 and then simplifying further it'd be .25(M)(v^2)? Am I on the right track?

Someone else I asked said that this does not even have rotational energy, so I'm very lost and I'd greatly appreciate any input! Thank you very much!

2. May 30, 2015

### Orodruin

Staff Emeritus
Yes.
They were wrong. Often, pulleys will be assumed to be ideal without mass and, consequently, with zero moment of inertia, leading to zero rotational energy. But if the pulley is massive and rotates with the string, then it will have rotational energy.

3. May 30, 2015

### haruspex

This is not at all clear.
The pulley is on a horizontal axis, yes?
Is the wooden block initially resting on the floor, more than .76m below, but the .2kg mass is dropped from table height? And the .2kg mass falls .76m before the string becomes taut? If not, I don't understand the rest of the question.
Since no mass or radius is given for the pulley, you have no way to take the pulley's rotational inertia into account, so you should probably treat it as massless. (But the question ought to make this clear.)

4. May 30, 2015

### Preptopro

I apologize, what I meant was that the pulled is hooked on the edge of a table and the wooden block is .8meters away from the edge of the table where the pulley is positioned, the wooden block is initially .8meters from the pulley and the .2kg mass is dropped from the table height or .76meters so when the .2kg mass is dropped the wooden block slides across the table .8 meters to the point of the pulley at a velocity of .92m/s. The radius of the pulley is .02meters and its mass is not known but assuming that I did know it, how would I go about finding the rotational energy then? and I apologize again, this is my first time dealing with pulleys.

5. May 30, 2015

### haruspex

Ok, that's clearer.
.92m/s is the speed when it reaches the pulley, yes? What else is moving at that time? (You should consider the moment before the mass reaches the floor.)
You do not seem to have executed the limear KE formula correctly.
If there is both frictional loss and rotational energy in the pulley then you do not have enough information.

6. May 30, 2015

### Preptopro

Yes correct .92m/s is the speed when the wooden block reaches the pulley.
The .2kg mass is also falling down at that time, do I need to calculate the the speed at which it falls? and if so how?

How can I correct this then?

I do not really have much prior experience with rotational kinematics or with pulleys really, and I've just started learning physics so I apologize if I seem really lost because I sort-of am

7. May 30, 2015

### haruspex

Yes. Is the string changing length during the fall? What does that tell you about the relative speeds of the two objects?
You correctly quoted the linear KE as $\frac 12 mv^2$, but you seem to have forgotten to square the velocity in the calculation.