Rotation with Newtons second law and work energy

[Jason03]

Heres the problem

http://img141.imageshack.us/img141/2820/49686394mt3.jpg

Now I have to work it out using Newtons second law and Work energy...

For Newtons second Law (F = ma)...would I be using

I = mk^2

M = I\alpha

than setting up F = ma...?


and for work energy I am using kinetic and potential...so the setup without using friction would just be using kinetic and potential?...and is the potential an elastic potential since its a wheel?

 

[konthelion]

Use \tau= I\alpha where \tau is the net torque external.
Let tension = T. Apply Newton's Second Law for rotational rotation T and angular acceleration

TR=I\alpha, where R is the radius of the pulley.

For the second part, Total Energy = Kinetic Energy + Potential Energy + Rotational Kinetic Energy
where Rotational Kinetic Energy = \frac_{1}{2}I\omega^2

 

[konthelion]

Sorry for double posting; I can't seem to edit my previous post, weird.

Rotational kinetic energy = \frac{1}{2}I\omega^2

 

[Jason03]

For this equ. that you had posted

TR=I\alpha

is that relating to F = MA ?

also the T in the equ is tension, correct?..and if so where does the \taucome in...

and just for clarification...my professor has been using the term "magnitude of the couple" instead of torque as you did...but the equ is the same...so I am assuming there one in the same...

 

[konthelion]

Yes, T is the tension. As for magnitude of the couple, I believe it is the same thing as torque. (I've never really seen that term used before)

In order to relate the suspended mass with Newton's 2nd law, you can draw a free diagram to see more cleary.

We know that,
\Sigma F_{net}=ma_{y}, m is the mass of the suspended object
\Rightarrow mg-T=ma_{t} Equation *, where T is the tension

But, we know that tangential acceleration a_{t}=R\alpha, R is the radius, alpha is the angular acceleration.
Then, \frac{mg-T}{m} = R\frac{TR}{I}
Then, T= \frac{mg}{1+(mR^2/I}
plug T back into Equation (*)

Solve for \alpha_{t} = \frac{1}{1+I/(mR^2)}
Plug back R, solve for angular acceleration. Then solve for angular velocity.
(Please note that this is for when there is no friction between the rope and the pulley!)

Now, for the second part, using conservation of momentum. Let the distance that the mass drop = d.

By Conservation of momentum,
U_{f}+K_{f} = U_i + K_f = 0 +0 = 0

Then, U_{f}=U_{object} + U_{pulley}=m_{object}g(-d)+m_{pulley}g(-d/2)
K_{f} = ... = \frac{1}{2}(m_{object}+m_{pulley})v^2

After some simplification, v=\sqrt{2m_{object}gd/(m_{object}+m_{pulley}}}
For d, plug in 5 revolutions from the problem. 5 revolutions are equal to 5*circumference of the pulley.
Then, solve for angular velocity.

Lastly, we know that \frac{1}{2}I_{pulley}\omega^2= \frac{1}{2}m_{pulley}v^2 by the rotational kinetic energy!

Hope that helped.

 

[D H]

Yes, T is the tension.

No, it is not. T (better written as the Greek letter tau, \tau) is torque, not tension. Tension is a force. Torque has units length*force. It is a different beast.

 

[konthelion]

No, it is not. T (better written as the Greek letter tau, \tau) is torque, not tension. Tension is a force. Torque has units length*force. It is a different beast.

Yes, I know. I just have a habit of writing tension as T and I believe I did use tau as torque and specified T as tension.

 

[Jason03]

My equ looks like this for the first section ..I just need to solve for Tension...

(also I replaced tau with M which is the magnitde of the couple, which is given in the problem)

\frac{mg-T}{m} = r\frac{M}{I}

also is the mass on each side of the equ. of the suspened object?

 

[konthelion]

Yes, the m=mass of the suspended object (both m's), same as m_{object} for the second method. mass of the pulley = m_{pulley}