Pulley angular velocity problem

[jeffrey]

Homework Statement

Suppose mass mL has a constant vertical speed v while hoisting, block B is stationary. In that case, calculate the rotational speeds ωQ, ωR and ωS of discs Q, R and S expressed in v and D. Define clockwise as positive

Relevant Equations

v=omega*radius
length_a+length_b= constant

Further given:
- every beam is infinite stiff
- pulleys are massless
- cables don't stretch, no slip, and frictionless.
-Every pulley has a diameter D except pulley Q. Pulley Q has diameter 0.5*D

So what I don't understand is how to calculate/determine the velocity at R and S. Can someone help me or send me into the right direction? I know from the answers that the velocity at R is 2v and at C is 1v, but I don't understand why.

 

[jambaugh]

I would start by relating the total length of the bottom cable to the height of the bottom platform. The rate of change of length of the cable will be the rate it is being taken up by $T_2$ and give you rotation rate of Q.

 

[haruspex]

Consider that each vertical section of string may have a different velocity.
What is the velocity of the leftmost? What about the one between S and R?

 

[jeffrey]

Consider that each vertical section of string may have a different velocity.
What is the velocity of the leftmost? What about the one between S and R?

First of all thank you for your response.
I'd say that the leftmost velocity is equal to 0, because that part is connected to the block and doesn't move.
But for the one between S and R I would guess that one is 2v, because it's kind of the same as a 'wheel without slip' situation. The center of mass of S moves with velocity v downward so the part between S and R is 2v.

 

[haruspex]

First of all thank you for your response.
I'd say that the leftmost velocity is equal to 0, because that part is connected to the block and doesn't move.
But for the one between S and R I would guess that one is 2v, because it's kind of the same as a 'wheel without slip' situation. The center of mass of S moves with velocity v downward so the part between S and R is 2v.

Yes, except that it says "hoisting", so C is rising.
So, how about the rest of the vertical sections?

 

[jeffrey]

Yes, except that it says "hoisting", so C is rising.
So, how about the rest of the vertical sections?

I'd say also 2v for all other vertical ropes, but my guess is based on intuition.

 

[haruspex]

I'd say also 2v for all other vertical ropes, but my guess is based on intuition.

Think through why it is that the rope on the left of pulley S is stationary but the one on its right is rising at 2v. Does the same principle apply at any other pulley?

 

[jeffrey]

Think through why it is that the rope on the left of pulley S is stationary but the one on its right is rising at 2v. Does the same principle apply at any other pulley?

Is it because pulley S performs a translational motion as well as a rotation?
Pulley R (and all other pulleys in the same row) on the other hand only perform a rotational motion and no tranlational motion. But because the rope between S and R is the same that rope moving with a velocity of 2v.
If the rope left of pulley R moves with 2v then the rope right of R should also move with 2v.
The rope between R and the pulley with the mass should also be 2v then, because it is the same piece of rope. But the pulley with the mass does move upwards so left of this pulley the rope has a speed of v and right of it the speed is 3v (I don't think that this part is correct ).

 

[haruspex]

the pulley with the mass does move upwards so left of this pulley the rope has a speed of v and right of it the speed is 3v

Closer... but it went from zero to 2v in going around pulley S, not from zero to v.
Think about the velocities of these three points: the leftmost point on S, the centre of S, and the rightmost point of S. What is the relationship?

 

[jeffrey]

Closer... but it went from zero to 2v in going around pulley S, not from zero to v.
Think about the velocities of these three points: the leftmost point on S, the centre of S, and the rightmost point of S. What is the relationship?

hmm do you mean v=omega*r?

 

[haruspex]

hmm do you mean v=omega*r?

No, I was asking about the three linear velocities and their relationship... but ωr is one way to connect them.

 

[jeffrey]

No, I was asking about the three linear velocities and their relationship... but ωr is one way to connect them.

ω*r gives the velocities at the left and right side of the pulley S, this gives a v upwards for the right side and v downward for the left side. This is the velocity by rotation, but the pulley also translates with a velocity v. This the translation v is for the left middle and right side the same. So, it adding the left side gives 0, for the middle v, and right 2v.

 

[haruspex]

ω*r gives the velocities at the left and right side of the pulley S, this gives a v upwards for the right side and v downward for the left side. This is the velocity by rotation, but the pulley also translates with a velocity v. This the translation v is for the left middle and right side the same. So, it adding the left side gives 0, for the middle v, and right 2v.

Good, so apply the same reasoning to the next pulley along.

 

[jeffrey]

Good, so apply the same reasoning to the next pulley along.

The next one doesn't translate only rotate, and because the right side of pulley S moved with 2v pulley R is rotating with 2v.

 

[haruspex]

The next one doesn't translate only rotate, and because the right side of pulley S moved with 2v pulley R is rotating with 2v.

I mean the next pulley along at the same level, the one the mass hangs from.

 

[jeffrey]

I mean the next pulley along at the same level, the one the mass hangs from.

Okay, that one rotates with a velocity of 2v but translates also upwards. So what I'd say is that the left side moves with 2v-v, the middle v and the right side 2v+v.

 

[haruspex]

rotates with a velocity of 2v

How fast is the string on its left moving down? If the pulley were not moving up, how fast would it be rotating? Does its upward movement change that?

 

[jeffrey]

How fast is the string on its left moving down? If the pulley were not moving up, how fast would it be rotating? Does its upward movement change that?

It is moving down with 2v right? So if it were not moving up it would rotate with 2v/r. I think the upward movement does change it. The upward movement gives the left, middle and right side a velocity of v upwards. While the rotation gives the left side a velocity of 2v downwards and 2v upward so adding the velocities does change it.
Or am i thinking wrong?

 

[haruspex]

The upward movement gives the left, middle and right side a velocity of v upwards. While the rotation gives the left side a velocity of 2v downwards and 2v upward

You lost me somewhere in there.
We know the left is moving down at 2v and the centre moving up at v. So what is the rotation rate, and what does that make the velocity of the right side?

 

[jeffrey]

You lost me somewhere in there.
We know the left is moving down at 2v and the centre moving up at v. So what is the rotation rate, and what does that make the velocity of the right side?

3v?

 

[jeffrey]

You lost me somewhere in there.
We know the left is moving down at 2v and the centre moving up at v. So what is the rotation rate, and what does that make the velocity of the right side?

I have an exam about this tomorrow could you please tell me?

 

[haruspex]

3v?

Not sure which question you are answering.
A rotation rate would be some constant multiplied by v/r, so I assume you have jumped straight to answering the velocity of the RHS. I ask intermediate questions for a reason.

If the rotation rate is $\omega$ then the left hand side is moving down at $r\omega$ relative to the centre. The centre is moving up at v, so the net velocity of the LHS is down at $r\omega-v$, but we know that it is moving down at 2v, so we have the equation $r\omega-v=2v$. This gives you the rotation rate.
Write the corresponding equation for the RHS.

 

[jeffrey]

Not sure which question you are answering.
A rotation rate would be some constant multiplied by v/r, so I assume you have jumped straight to answering the velocity of the RHS. I ask intermediate questions for a reason.

If the rotation rate is $\omega$ then the left hand side is moving down at $r\omega$ relative to the centre. The centre is moving up at v, so the net velocity of the LHS is down at $r\omega-v$, but we know that it is moving down at 2v, so we have the equation $r\omega-v=2v$. This gives you the rotation rate.
Write the corresponding equation for the RHS.

Thank you a lot!
RHS would be $r\omega+v=4v$, with r*omega being 3v right.

 

[haruspex]

Thank you a lot!
RHS would be $r\omega+v=4v$, with r*omega being 3v right.

Yes, so now you can see the pattern: each turn around a lower pulley adds 2v.

 

[jeffrey]

Yes, so now you can see the pattern: each turn around a lower pulley adds 2v.

ahaa i get it now thank you a lot!