How Does the Velocity of a Falling Disc's Centre Change with Height h?

[mooncrater]

question - Consider the cobination of two discs(attached) , then what will be the velocity of falling disc centre as a function of h. Both discs are identical and string doesn't slip relative to disc.

Relevant equations -
Conservation of energy .
Torque=r×F
Force=ma

An attempt to the question -
Attached
When I saw this question for the first time , I tried doing it with the force and torque method through which I got the acceleration and calculated the velocity of the lower disc.
But When I saw its solution, which was done by the conservation of energy , also seemed correct to me, but I didnt understand what was wrong with my method (duh).
So can someone please point out my mistake , I would be very thankful for that.

 

[paisiello2]

Maybe you could state what the question was?

 

[TSny]

I believe your method has the correct answer. I'm not sure I follow why you defined $\beta_{net} = \beta_1 +\beta_2$. There is nothing rotating at the rate $\beta_{net}$. But it is true that the linear acceleration of the center of the lower disk is $a = (\beta_1 +\beta_2)r$.

I think there is a mistake in the energy approach. $v = \omega r$ is not valid.

 

[Vatsal Sanjay]

$v$ is not equal to $rw$ I think what you are missing here is the constraint relationship.

 

[TSny]

Mooncrater,
Note that even if the lower disk did not rotate, the center of the lower disk would have a velocity due to the rotation of the upper disk. So, $v$ for the lower disk depends on both the rotation of the upper disk as well as the rotation of the lower disk.

 

[Vatsal Sanjay]

Note that even if the lower disk did not rotate, the center of the lower disk would have a velocity due to the rotation of the upper disk. So, vv for the lower disk depends on both the rotation of the upper disk as well as the rotation of the lower disk.

I meant exactly the same. Right? (*sorry but I could not get if anything is wrong about my statement)

 

[TSny]

I meant exactly the same. Right? (*sorry but I could not get if anything is wrong about my statement)

Yes, you are right. We are in agreement.

 

[mooncrater]

I think there is a mistake in the energy approach. $v = \omega r$ is not valid.

But the thread doesn't SLIP over the lower disc so how is υ=rω not valid here ?Is it due to the fact that the velocity of the lower disc is not only dependent on the angular velocity of the lower disc but also of the upper disc ( as u said )?But then what will be relation between v and w (omega)?

 

[Vatsal Sanjay]

But the thread doesn't SLIP over the lower disc so how is υ=rω not valid here ?

Yes. Exactly. The thread doesn't slip over the disc. However the speed of thread is not zero. if $v$ is the velocity of center of mass and $\omega_2$, the angular velocity of lower disc. Then the velocity of point on disc in contact with the thread will be $v - r\omega_2$ (downwards). For non slip condition this quantity must be equal to the velocity of the thread at the point of contact. Had that been zero, then your relation was correct. I guess I am not supposed to write the entire expression. So I will leave it up to you to figure out what the relation is. :)

 

[mooncrater]

So then the relation should be:
ν₂-ω₂r=ω₁r
Am I correct?
Where ω₁ is the angular velocity of the upper disc..
Edit :sorry instead of ω₂ it was ω₁

 

[Vatsal Sanjay]

ν2-ω2r=ω1r

Where ω2 is the angular velocity of the upper disc..

It is correct considering that you have written it in the frame of reference of the centre of mass of the lower disc,
If the above equation is written in the ground (inertial) frame of reference, then you have interchanged $\omega_1$ and $\omega_2$ .
See the relation is same in both the cases but writing the above equation in ground frame is technically incorrect. (Mathematical manipulations apart)

I hope I am clear with my point?

 

[mooncrater]

Is it correct after I edited it?

 

[Vatsal Sanjay]

Yes.. Its correct now!

 

[mooncrater]

Thank you...now its clear to me...