Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A pulley having some mass has a massless string around it, with two unequal masses at the end of the string. Sufficient friction is present between pulley and the string such that the string does not slip on the pulley .

The tensions in the massless string on the two sides of pulley will be same or different?

According to me the tensions on the two sides should be different so as to provide a net torque to the pulley .But what I have learned till now is that the tension in a massless string is same throughout.

Let the tensions on the two sides of the string be $T_1$ and $T_2$.

Writing equation for rotational motion of pulley we have (T1-T2)R = Ia

Since the pulley has non zero moment of inertia and angular acceleration ,T1 WILL NOT BE EQUAL TO T2.

I am having difficulty in understanding that how can a massless string have different tensions.

share|cite|improve this question

closed as off-topic by centralcharge, jinawee, Waffle's Crazy Peanut, Qmechanic Dec 19 '13 at 14:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – centralcharge, jinawee, Waffle's Crazy Peanut, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

Usually, to simplify things, the pulleys are assumed to be massless/frictionless.

In this case, if the tensions were different. we would have a net torque. However, I=0 so this is impossible without a persistent infinite angular acceleration (which is also not possible). So we take the tensions to be the same.

Similarly, in the frictionless case, there is no way for the string to transmit torque, and different tensions must result in a nonzero torque, so the tensions can't be different.

You are correct in saying that a pulley with mass will make the string have different tensions. In this case, we can use the net force on the pulley and its angular acceleration (which can be calculated from constraints) to form another equation to account for the extra variable introduced.

share|cite|improve this answer
@Vibhor: Why do you think that a massless string cannot have different tensions? There's no reason for it -- when the string is obstruction-free this happens because otherwise parts of the string would fly off to infinity. But in the case of a pulley, the pulley is there to equalize the imbalance. – Manishearth Jun 13 '13 at 8:05
@Vibhor: If the tension was uneven, parts of the string would have a net force. It's massless, so the net force gives infinite acceleration. – Manishearth Jun 13 '13 at 9:22
@Vibhor: Well, the pulley needs mass too. When the pulley has mass and friction, the force imbalance results in a net torque on the pulley (it cannot do so if the pulley is massless or frictionless), and is taken care of. – Manishearth Jun 13 '13 at 10:10
@Vibhor: Go up three comments. I checked your work, and told you what to do next. Thing is, if I keep doing that, you'll never understand the concept here, not properly. Basically I was asking you to experiment with the system. It's not entirely a "fixed path" where I can check which track you're on. You will understand the concept automatically if you do this for a while. (Also, I remember being in your shoes a few years ago; and experimenting with the system is where I got my conceptual understanding from) – Manishearth Jun 14 '13 at 12:01
There's a big difference between helping and spoonfeeding. I didn't want to do the latter; as in the long run it doesn't help. – Manishearth Jun 14 '13 at 12:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.