# One more question(hopefully) before exam

1. Dec 10, 2003

### moooocow

Two blocks are connected by a string of negligible mass passing over a pulley of radius .250m and moment of inertia I. The block on the frictionless incline(it has a picture of incline with block 1 on it at theta = 37, and block 2 is hanging off the side by the pulley) is moving up with a constant acceleration of 2 m/s^2. I can find the moment of inertia of the pulley if I have the tensions T1(block 1 before the pulley) and T2(block 2 after the pulley) I am having trouble with 2 things. Why are the tensions different in the two parts of the string and are the tensions just T2 = (m2)a + (m2)g and T1 = (m1)a - (m1)gsin(theta) with a = 2? Because these do not give me the correct tensions, but I am thrown off in the first place by the tensions being different, any help would be VERY appreciated. Thank you very much.

2. Dec 10, 2003

### moooocow

And if anyone could help with this one I would be forever in debt to you, hehe.

A constant horizontal force is applied to a lawn roller in the form of a uniform solid cylinder of raidus R and mass M. Show that the acceleration of the center of mass is 2F/3m. the minimum coefficient of friction neccesary to prevent slipping is F/3mg. If anyone could just give me a quick overview on how to approach these, that would be great. Thanks alot for any help

3. Dec 10, 2003

### himanshu121

Where the Force F is acting

4. Dec 10, 2003

### moooocow

F basically splits up and connects on both sides of the roller in the center of the end caps. At first I thought you could use torque to find the angular acceleration and use that to find the center of mass acceleration but my answers are not comming up correct.

5. Dec 10, 2003

### himanshu121

How you have applied torque equation if the force is acting at the centre. And is it starting rolling without sliding

6. Dec 10, 2003

### moooocow

Well, that would be why it isnt working, Im not really sure how to go about it, besides the kinetic energy, but im not sure exactly how that would work with it just rolling horizontally.

7. Dec 10, 2003

### himanshu121

i got the answer just wait

8. Dec 10, 2003

### himanshu121

Here

Applying the force equation
$$F-f=Ma_{cm}$$$$//f=\frac{MR^2\alpha}{2}$$
now as there is rolling without slipping
$$a_{cm}=r\alpha$$
solving the above equation
u will get $$a_{cm}=\frac{2F}{3M}$$

Last edited: Dec 10, 2003
9. Dec 10, 2003

### himanshu121

Where f is frictional force

which will be equal to $$\mu Mg=f$$
where f=F/3

Last edited: Dec 10, 2003