Master Dynamics: Answers to Practice Questions and Concept Clarification

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The discussion focuses on solving physics problems involving pulleys and masses, specifically addressing the acceleration of pulleys and tension in strings. Participants highlight the importance of applying Newton's second law to each mass and pulley, noting that the tension differs on either side of a pulley due to its mass. There is confusion about the assumption of massless pulleys and the implications for torque and tension. The conversation also questions the clarity of the problem, suggesting that the phrasing might mislead readers into expecting rotational motion rather than frictionless sliding. Overall, the thread emphasizes the need for careful analysis of forces and assumptions in physics problems.
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fterh said:
Some background information: I'm doing some reading up for PhO (so it's beyond what I'm supposed to learn), and so I'll post all my questions here (regarding both concepts and actual practice questions). Sorry if you feel there is a lack of effort on my part, but sometimes I'm really lost and don't know where to start.

1) A light inextensible string passes round a fixed smooth pulley and carries at each end a smooth pulley of mass 1.0 kg. Over each of these pulleys a string hangs, one carrying masses of 1.0 kg and 2.0 kg at its ends, the other masses of 1.0 kg and 3.0 kg. The system is set in motion. Find the acceleration of the pulleys and the tension in the string to which they are attached.

The answer is 0.426 m s^-2, but I just couldn't get it! Ugh.

2) Two steel balls of masses M and m are suspended by vertical strings so as to be just in contact wit their centers at the same height. The ball of mass M is pulled to one side, keeping its center in the vertical plane which originally contained the centers. It is released from rest when its height is h above the original position.

a) Show that, whatever the value of m, the second ball cannot rise to a height above its equilibrium position greater than 4h.

No idea. A clue, please? :)
 
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The logical place to start on the first question would be to write Newton's second law for every mass and the two pulleys. Did you try that?
 
Pi-Bond said:
The logical place to start on the first question would be to write Newton's second law for every mass and the two pulleys. Did you try that?

Just checking, since the 1kg pulleys are not massless, so the tension of the string on either side of the pulley is different?
 
Yes, the tension would be different on both sides. You will have to write Newton's second law for both pulleys like normal.
 
Pi-Bond said:
Yes, the tension would be different on both sides. You will have to write Newton's second law for both pulleys like normal.

How do I proceed?

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Since the pulleys are not massless, they will not have zero accelerations. Also, the tension in a thread is the same throughout, so the tensions T1 and T2 are equal, and T3 and T4 are equal. What I meant in the previous post was that the tension of the thread connecting the pulleys will not be equal to the tensions of the threads below them. Write Newton's laws for the pulleys in terms of the tensions. Then write Newton's laws for every mass, keeping in mind that the masses accelerate along with the pulleys they are connected to.
 
Pi-Bond said:
Since the pulleys are not massless, they will not have zero accelerations. Also, the tension in a thread is the same throughout, so the tensions T1 and T2 are equal, and T3 and T4 are equal. What I meant in the previous post was that the tension of the thread connecting the pulleys will not be equal to the tensions of the threads below them. Write Newton's laws for the pulleys in terms of the tensions. Then write Newton's laws for every mass, keeping in mind that the masses accelerate along with the pulleys they are connected to.

Why is the tension in a thread the same throughout? If the pulley has a mass, a resultant torque is required to set it in rotational motion, thus the tension on one side has to be greater than the other side, no?
 
That is correct, but you have no information of the radius of the pulleys, which is required to write a complete equation for torque. If you "ignore" the rotation of the pulley, you do end up with the answer you mentioned in your post. I'm not sure why you can do this, but I guess the two lower pulleys are not free to rotate about their central axes, so only the threads slide across them without friction (thus having the same tension)
 
Pi-Bond said:
That is correct, but you have no information of the radius of the pulleys, which is required to write a complete equation for torque. If you "ignore" the rotation of the pulley, you do end up with the answer you mentioned in your post. I'm not sure why you can do this, but I guess the two lower pulleys are not free to rotate about their central axes, so only the threads slide across them without friction (thus having the same tension)

Is this a flaw in the question? Because when it says "pulley", it's only natural to assume that rotational motion is involved, rather than frictionless sliding.
 
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It would definitely appear so...where did you get the question from by the way?
 
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