0
$\begingroup$

A string is wrapped several times on a cylinder of mass $M$ and radius $R$. The cylinder is pivoted about its axis of symmetry.

A block of mass $m$ is tied to the string and it rests on a support so that the string doesn't slack. The block is lifted vertically at a distance$h$ and the support is removed. The question is to find the velocity of the block and the angular velocity of the cylinder just after the string is taut.

In its solution, angular momentum conservation is applied for the reason that tension is impulsive so the force due to gravity can be neglected. Moreover, the tension in the string, being an internal force, gets canceled, so the net torque on the system is zero.

I'm unable to understand that in most of the cases we do consider the torque due to tension, then why don't we do it the same way here as well? Is it because the tension being impulsive? If it is so then what's the exact reason behind this concept?

$\endgroup$

1 Answer 1

0
$\begingroup$

The question says 'just after' the string goes taut.

The impulse to both the block and the cylinder when the string goes taut is a sudden high force over a short time - it causes the change in momentum of the block and the change in angular momentum of the cylinder. This kind of question uses conservation of momentum.

It's true that the force of gravity is used to determine what the speed of the block would be, when the string goes taut, but not needed again at the time when conservation of momentum is applied.

Different questions where the tension in the string is constant and acts over a longer time often use 'equations of motion' that are derived assuming that the acceleration or angular acceleration is constant.


The tension in the string at the moment when the string goes tight, causes the change in momentum for both objects (of equal magnitude) as follows, there is a varying force $F$

From $F=ma$

$F=m\frac{dv}{dt}$

integrating

$\int F dt = \int m dv$

so the impulse equals the change in momentum

$\endgroup$
2
  • $\begingroup$ My exact question was that , why the torque due to tension has not been considered here? Is it because the tension is impulsive? If so, what's the reason for this? $\endgroup$
    – Aspirant
    May 31, 2021 at 8:44
  • $\begingroup$ Ok, yes technically, the torque is due to the tension in the string, it's just a different way (to other questions) of getting the result. The resultant change in angular momentum of the cylinder, due to the suddenly increasing tension is the integral of torque times time and at any moment torque is tension times radius. It's just that since it varies suddenly we can't use the equations of motion, (those need constant acceleration), so conservation of momentum is used instead. That method appears to miss out torque and tension, although it is how the impulse is transmitted (answer modified) $\endgroup$ May 31, 2021 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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