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If you drop a rod with a small mass on one end and a large mass on the other (initially parallel to the ground) does it rotate as it falls?

If you take the pivot point to be the center of mass, then we should expect no rotation. But if you take the pivot point to be one end of the rod, then we expect rotation. I've confused myself here, as that can't be the case.

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To compute the torque on a body in free fall, you have to use the distance of each force to the center of mass. When you offset your point of reference, you need to account for that by subtracting the force of gravity on the center of mass multiplied by the distance of the offset. When you do that, the rod will continue to be stationary, as expected.

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In the absence of air resistance, the assembly won't rotate if no rotational motion was imparted when it was released. If you consider a free-body diagram of the assembly in a (non-inertial) reference frame moving with it, then there will be 'virtual' d'Alembert forces, due to the fact that acceleration is present, which will exactly balance the weight of each body. Because of this, there is no net force on any of the bodies in that reference frame and there will be no net torque regardless of which point you choose to sum the torques about.

As you will see from the linked article, this d'Alembert force is closely related to the coriolis and centrifugal forces, which are other virtual forces that are used in similar non-inertial reference frames.

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    $\begingroup$ D'Alembert force- that's a new word I learnt today. Thanks. $\endgroup$
    – Floris
    Jan 14, 2015 at 1:31
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    $\begingroup$ @Floris: No problem - I just remembered it from Mechanics classes at uni :-) $\endgroup$
    – Time4Tea
    Jan 14, 2015 at 2:03
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The rod doesnot rotate because 1) both the mases (heavier mass B and smaller mass A on both ends )move down with same acceleration,so they dont rotate. 2)and you asked the question if you take the pivot point at centre of mass there is no torque and if you take other end as pivot then you got confused ,but the other pivot point (A) is not stationary and accelerating so its in non inertial frame,so if you calculate in non inertial frame then you should add pseudo force in the opposite direction at mass B so you get again resultant torque=0. I think your doubt is solved.enter image description here

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But if you take the pivot point to be one end of the rod, then we expect rotation.

If you take the pivot point to be one end of the rod, you will find that there is a net torque on the rod at that point. $\tau_{net} > 0$. You might expect that means that there is rotation (or at least that there is a change in the rotation). But that isn't the case.

In linear motion we equate a net force with a change in momentum. In this case we can equate a net torque with a change in angular momentum. When the center of mass is not accelerating, there must be a change in angular velocity. But in the case of your falling rod, there isn't one.

Instead, we can calculate the angular momentum of the rod about the same axis. The angular momentum of a point mass is $L = m v d$. For $m1$, the distance to the axis is zero, and it has no angular momentum. But $m2$ is located $d$ distance away, and it has downward velocity. In freefall, this velocity is changing, so the total angular momentum of the system is changing as well.

The net torque on the end of the rod is describing a system with a changing angular momentum. It's just that it is not associated with a changing angular velocity.

If you calculate the torque about the center of mass, then the contribution of all the masses cancel out and any change in angular momentum will be associated with a change in angular velocity.

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  • $\begingroup$ Interesting perspective - change in angular momentum without a change in angular velocity. I get the concept by I'm not sure it clarifies things for OP. $\endgroup$
    – Floris
    Jan 14, 2015 at 2:52

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