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Problem:

A block tied to a string is rotating with an angular velocity $\omega$ on a frictionless table. The string passes through a hole in the center of the table. If the string is pulled and the length of the string reduces to half, find the new angular velocity of the block.

Question:

I know this question can be easily solved using conservation of angular momentum of the block because no external torque acts on it. The answer to this question will be $4 \omega$.

What I wanted to know was that why would the the speed change at all? The tension due to string acts perpendicular to the direction of motion so it cannot change the angular velocity. So which force changed the angular velocity of the block??

marked as duplicate by sammy gerbil, stafusa, John Rennie newtonian-mechanics Sep 18 at 5:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 6
    If the velocity were always tangential the block would just go in a circle. For the block to spiral inwards the velocity must point slightly inwards. That means the tension and direction of motion are not perpendicular. – John Rennie Sep 14 at 11:30
  • @JohnRennie Your answer does seem correct. Thanks. – Harshit Joshi Sep 14 at 11:44
  • @JohnRennie. You should make that comment an answer. – md2perpe Sep 14 at 12:52
  • Even if the block were to jump instantaneously to a point half as far from the center, keeping its linear velocity unchanged, the angular velocity would still jump, although only to $2\omega$ -- because a radian is now a shorter distance for it. The other factor of $2$ comes from increasing the kinetic energy as described in the accepted answer. – Henning Makholm Sep 14 at 16:31
up vote 2 down vote accepted

As the block rotates it is pulled inward via the tension in the string.

In doing so the block must move towards the hole at position $O$ which is the same direction as that of the tension $\vec T$ so there must be work done $\vec F \cdot d\vec r$ by the force which results in an increase in the kinetic energy of the block.

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As you have pointed out the torque exerted by the tension $\vec R \times \vec T=0$ and so the angular momentum is a conserved quantity.

When the block is rotating in a circle of radius $r$ it has a rotational kinetic energy (considering potential energy to be 0 on the table) given by $$\dfrac{1}{2}I\omega ²$$

According to conservation of energy, energy can neither be created nor be destroyed (forget about Einstein). The block should possess an energy equal to $$\dfrac{1}{2}I\omega ² + \Delta W$$ What is $\Delta W$ then. It is the work done by the tension on the block as it moves inwards in a circle of radius $\dfrac{r}{2}$.

If the block were to move with the same angular velocity it would mean loss in energy (remember conservation of energy). The block is bound to move with a higher angular velocity.

In the similar way if no external torque acts on a body it's angular momentum will be conserved (otherwise energy will be destroyed).

Velocity of an object is not due to an external force but due to it's energy. If a body at rest it set into motion by applying an impulse the body will move with a velocity $v$, because the body has been given some energy.

Hope it might help!

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