Finding angular velocity using conservation of energy

[Coderhk]

Homework Statement

In the document below I need to try and find the angular velocity
I need help on part b

Homework Equations

F=ma
KE_Translational= 0.5mv^2
KE-Rotational= 0.5Iw^2
Assuming g=10m/s^2

The Attempt at a Solution

I have the answer key attached with the question but I am not sure why the answer key uses 2.5J as the sum of the kinetic energy. Why wouldn't the energy be 6.25J instead. 6.25J is the total gravitational potentia; energy relative to the ground. So shouldn't all that energy be converted into kinetic energy?

 

[ehild]

Homework Statement

In the document below I need to try and find the angular velocity

Homework Equations

F=ma
KE_Translational= 0.5mv^2
KE-Rotational= 0.5Iw^2
Assuming g=10m/s^2

The Attempt at a Solution

I have the answer key attached with the question but I am not sure why the answer key uses 2.5J as the sum of the kinetic energy. Why wouldn't the energy be 6.25J instead. 6.25J is the total gravitational potentia; energy relative to the ground. So shouldn't all that energy be converted into kinetic energy?

The problem says "Calculate the total kinetic energy of the cylinder when it reaches the table".

 

[Coderhk]

The problem says "Calculate the total kinetic energy of the cylinder when it reaches the table".

I need help on part b not a

 

[TSny]

Does the angular velocity of the cylinder change after the cylinder leaves the table?

 

[Coderhk]

Does the angular velocity of the cylinder change after the cylinder leaves the table?

No, since there is no net external torque

 

[TSny]

No, since there is no net external torque

Right. Does that help to see why they used 2.5 J to get the answer for part b?

 

[Coderhk]

Right. Does that help to see why they used 2.5 J to get the answer for part b?

I see since the angular velocity at the table is the same as at the floor, we just calculate the angular velocity at the table. But why can't we attempt to find it at the floor directly by saying all the gravitational potential energy relative to the floor becomes kinetic energy?

 

[TSny]

All of the gravitational PE relative to the floor does become KE. But, that alone doesn't tell you what fraction of the KE at the floor will be rotational KE.

 

[Coderhk]

All of the gravitational PE relative to the floor does become KE. But, that alone doesn't tell you what fraction of the KE at the floor will be rotational KE.

Ahh, So does that mean that it wouldn't work because the transnational kinetic energy follows a trajectory rather than a straight line thus we can't say it is 0.5mv^2?

 

[TSny]

Ahh, So does that mean that it wouldn't work because the transnational kinetic energy follows a trajectory rather than a straight line thus we can't say it is 0.5mv^2?

I don't think that's the crucial point. In the solution, they use KE = ##\frac{3}{4}##M(Rω)². Under what condition is this true? Is this condition met once the cylinder leaves the table?

 

[Coderhk]

I don't think that's the crucial point. In the solution, they use KE = ##\frac{3}{4}## M(Rω)². Under what condition is this true? Is this condition met once the cylinder leaves the table?

That can only be achieved if they assume the ball is rolling without slipping which implies v=rw. Would it be that it is no longer true because the bottom of the ball is no longer not moving with respect to the ground?

 

[TSny]

d

That can only be achieved if they assume the ball is rolling without slipping which implies v=rw.

Yes.

Would it be that it is no longer true because the bottom of the ball is no longer not moving with respect to the ground?

I think you have it. Once the cylinder leaves the table it is no longer rolling without slipping. v increases while ω remains constant. So, as you say, v ≠ Rω after the cylinder leaves the table.

 

[Coderhk]

d Yes. I think you have it. Once the cylinder leaves the table it is no longer rolling without slipping. v increases while ω remains constant. So, as you say, v ≠ Rω after the cylinder leaves the table.

I see, thankyou very much. I get it now.

 

[TSny]

OK. Good work.