One more question(hopefully) before exam

[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.

 

[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 a lot for any help

 

[himanshu121]

Where the Force F is acting

 

[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.

 

[himanshu121]

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

 

[moooocow]

Well, that would be why it isn't working, I am not really sure how to go about it, besides the kinetic energy, but I am not sure exactly how that would work with it just rolling horizontally.

 

[himanshu121]

i got the answer just wait

 

[himanshu121]

Here

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

 

[himanshu121]

Where f is frictional force

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