Rotational motion and moment of inertia-HELP

[eyeshiled21]

rotational motion and moment of inertia---HELP!

Homework Statement

Two masses m1=2.00 kg and m2=1.00 kg are linked by a thread and thrown over a pulley whose mass is 1.00 kg. The radius of the pulley is 0.200 m.

a) Find the acceleration a with which the weights move
b) the tensions T1 and T2 of the threads which the masses are attached to.

Consider the pulley as a homogeneous disk. Disregard friction.

The Attempt at a Solution

Well, i tried to use the energy method because i know how to do that. But I'm having a problem because the problem doesn't state any given energy. please help me

 

[AlchemistK]

The energy method won't be very useful here, since they haven't asked for the velocity after the masses change their height.
Try making a free body diagram and work it out by its dynamics.

 

[eyeshiled21]

so you're saying that i must use Newtonian approach to solve this problem? well, I've thought of that at first. But then again, there's this moment of inertia of the pulley, and it's stated there that the pulley is a homogeneous disk. Am i really going to ignore that moment-of-inertia thing? because i think it will be useful

 

[AlchemistK]

No you don't have to ignore the moment of inertia. Find out the angular acceleration of the pulley, due to the torques by the two tensions. Now how will this angular acceleration be related to the acceleration of the thread?

 

[eyeshiled21]

will you help me using the torque method then? because what the instructor taught us was just up to moment of inertia.

 

[AlchemistK]

T= F * R sinθ

Where T is torque, F is force, and θ is the angle between the object and the force vector (90 degrees in this case).

And now just like F=ma, T= I * α

Where I is the moment of inertia and α is the angular acceleration.

Now since the thread is not sliding, and the pulley is purely rolling a= R*α (This has an important concept and proof behind it, make sure that you go through it, yourself, or with help of your teacher).

No simply make the free body diagrams of all the 3 objects and use the given formulas.

(And just like in dynamics, acceleration= net force/ total mass, here too, its the same thing, and to account for the pulley's mass, we use I/(R^2), this is just a shortcut, and might not give the correct result always since it has loopholes. In your example, a= (M1g-M2g)/[M1+M2+ Mp/2]. Where Mp/2 is Inertia of pulley upon R^2)

 

[eyeshiled21]

gee, thanks a lot...maybe i will show you my solutions tomorrow...

 

[eyeshiled21]

btw, is that a you give in your reply the linear acceleration or the angular?? because if that's linear, then that will be my answer in a.) right?

and also, is my formula for t1 and t2 correct?
t1=m1g-m1a and t2=m2a+m2g

 

[AlchemistK]

btw, is that a you give in your reply the linear acceleration or the angular??

the α is angular acceleration, and since there is no slipping a(linear acc.) = R*α

and also, is my formula for t1 and t2 correct?
t1=m1g-m1a and t2=m2a+m2g

Yes that's right.

 

[eyeshiled21]

when i multiply the angular acceleration by the radius of the pulley, my units are m^2/s^2 which for me, looks the wrong unit for acceleration. where did i get wrong?