Finding the angular velocity and rotational kinetic energy

[PhyIsOhSoHard]

Homework Statement

A solid cylinder is rolling along a horizontal plane and is friction less around its symmetric axis. The cylinder is pulled by a constant force, F and travels the distance, d. The cylinder does not glide and has a friction force, f on the ground.

Known values:
Mass: M
Radius: R
Moment of inertia: $I=\frac{1}{2}MR^2$
Force: F
Angle of the force: θ
Friction force: f
Distance traveled: d
Angular acceleration: α

Find the angular velocity and rotational kinetic energy for the cylinder when it starts at rest and travels the distance d.

Homework Equations

\frac{d\omega}{dt}=\alpha

K=\frac{1}{2}I\omega^2

The Attempt at a Solution

From a previous question, I found the angular acceleration:

I know the relation between angular velocity and angular acceleration is:
\frac{d\omega}{dt}=\alpha

However what got me confused is the time, t. If I can integrate my angular acceleration in terms of t then I can find my angular velocity but I don't know the time?

 

[haruspex]

Any conservation law you might be able to use?

 

[PhyIsOhSoHard]

Any conservation law you might be able to use?

I really have no clue. Maybe energy conservation between the state at rest and when it has traveled d distance? I'm not sure...

 

[haruspex]

I really have no clue. Maybe energy conservation between the state at rest and when it has traveled d distance? I'm not sure...

Sounds appropriate. What work will F have done on the system?

 

[PhyIsOhSoHard]

Sounds appropriate. What work will F have done on the system?

It would have done:
W=F\cdot \cos(\theta)\cdot d

Right? I'm not sure how to use this in my energy conservation equation.

 

[Gurdian]

Why would it even have angular velocity if the force is being applied through its cm?

 

[PhyIsOhSoHard]

Why would it even have angular velocity if the force is being applied through its cm?

Because I also have a friction force.

 

[PhyIsOhSoHard]

Anyone?

 

[haruspex]

It would have done:
W=F\cdot \cos(\theta)\cdot d

Yes.

I'm not sure how to use this in my energy conservation equation.

Where has that work gone? It's rolling contact with the ground, and there's no friction at the axle, so no work has been done overcoming friction.

 

[Gurdian]

Because I also have a friction force.

sorry I thought it was a frictionless surface.

 

[PhyIsOhSoHard]

Yes.


Where has that work gone? It's rolling contact with the ground, and there's no friction at the axle, so no work has been done overcoming friction.

Has it gone to the friction between the ground and the cylinder?

 

[PhyIsOhSoHard]

I'm still not sure about this.

 

[haruspex]

Has it gone to the friction between the ground and the cylinder?

No, as I posted before, that's rolling contact, so the friction does no net work. The work energy is still in the system. What form is it taking? You have a choice of KE and/or PE.

 

[PhyIsOhSoHard]

No, as I posted before, that's rolling contact, so the friction does no net work. The work energy is still in the system. What form is it taking? You have a choice of KE and/or PE.

I'd say there is KE. We have KE because the cylinder has a velocity but we have zero PE because our plane is horizontal.

 

[PhyIsOhSoHard]

Since I have no PE and the energy is conserved I have:
K_1=K_2
\frac{1}{2}m\cdot 0=\frac{1}{2}mv^2

Because it starts at rest so the velocity is zero.

I also know that the angular velocity is:
v=R\omega

So if I can find an expression for the velocity after it has traveled distance d then I can find my angular velocity as well. Problem is how do I find the velocity...

 

[haruspex]

Since I have no PE and the energy is conserved I have:
K_1=K_2
\frac{1}{2}m\cdot 0=\frac{1}{2}mv^2

Because it starts at rest so the velocity is zero.

I also know that the angular velocity is:
v=R\omega

So if I can find an expression for the velocity after it has traveled distance d then I can find my angular velocity as well. Problem is how do I find the velocity...

Suppose it has velocity v. Write out expressions for the angular velocity, the linear KE, the rotational KE and the total KE.

 

[PhyIsOhSoHard]

\omega=\frac{v}{R}

KE_{linear}=\frac{1}{2}mv^2

KE_{rotational}=\frac{1}{2}I\omega^2

KE_{total}=KE_{linear}+KE_{rotational}=\frac{1}{2}mv^2+\frac{1}{2}I \omega^2

 

[PhyIsOhSoHard]

Suppose it has velocity v. Write out expressions for the angular velocity, the linear KE, the rotational KE and the total KE.

I'm not sure how it will help me.

 

[haruspex]

\omega=\frac{v}{R}

KE_{linear}=\frac{1}{2}mv^2

KE_{rotational}=\frac{1}{2}I\omega^2

KE_{total}=KE_{linear}+KE_{rotational}=\frac{1}{2}mv^2+\frac{1}{2}I \omega^2

Good. You know how to write ω in terms of v and R, and you know how to write I in terms of M and R. So you can get KE in terms of M, v and R.
You previously wrote

K₁=K₂

but that's wrong. That would be true if no work had been done on the system. The force F has done work on the system, and you already calculated that value. So what equation should you write for work conservation?