Rotation with Newtons second law and work energy

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

[tex]TR=I\alpha[/tex], 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 = [tex]\frac_{1}{2}I\omega^2[/tex]

 

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

Rotational kinetic energy = [tex]\frac{1}{2}I\omega^2[/tex]

 

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,
[tex]\Sigma F_{net}=ma_{y}[/tex], m is the mass of the suspended object
[tex]\Rightarrow mg-T=ma_{t}[/tex] Equation *, where T is the tension

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

Solve for [tex]\alpha_{t} = \frac{1}{1+I/(mR^2)}[/tex]
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,
[tex]U_{f}+K_{f} = U_i + K_f = 0 +0 = 0[/tex]

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

After some simplification, [tex]v=\sqrt{2m_{object}gd/(m_{object}+m_{pulley}}}[/tex]
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 [tex]\frac{1}{2}I_{pulley}\omega^2= \frac{1}{2}m_{pulley}v^2[/tex] by the rotational kinetic energy!

Hope that helped.

 

No, it is not. T (better written as the Greek letter tau, [itex]\tau[/itex]) 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.

 

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