How can I make sense of the work-energy theorem in the situation of a force applied on a disk?

Question

Pretend that we are in space and that gravitational attraction is negligible. Now, imagine that there are two disks that are exactly the same. However, on the first disk, a force is applied tangentially and on the second disk, the same force is applied to its center of mass. The force is exerted over the same distance for both.

Now here is where I'm having problems making sense of the situation. Since the force is the same for both disks the horizontal acceleration should be the same. After travelling a distance $d$, both disks should have the same final horizontal velocity. The work done should also be the same.

However, since the force is tangential on the first disks, there must be a torque and the first disk would have gained an angular velocity. But $W = Δk_{rotation} + Δk_{translation}$. The work and the kinetic energy of translation can't be the same if disk 1 has rotational kinetic energy.

One of the assumptions I have made must therefore be wrong. Either the work done on the disks isn't the same, or the horizontal acceleration isn't the same, or disk 1 has no angular velocity.

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Pierre

Comments

physics.stackexchange.com/q/593279/267970 possible duplicate

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Prateek Mourya

Answer 1

However, on the first disk, a force is applied tangentially and on the second disk, the same force is applied to its center of mass. The force is exerted over the same distance for both.

It is not possible for the force to be applied for both the same amount of time and over the same distance. The tangentially applied force will cover a larger distance over the same time, or conversely will cover the same distance in a shorter time.

If you match time then their ending momentum will be the same, but the tangential disk will have higher energy corresponding to the rotational energy and the additional distance over which the force was applied.

If you match the distance that the material at the point of contact with the force moves (this is not the same as the distance that the CoM moves for the tangential disk) then the tangential force will be applied only briefly. The tangential disk will have less momentum but the same energy. That energy will be split between center of mass KE and rotational energy.

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Dale

Comments

To add to @Dale's answer. To apply a steady force tangent to the disk you need to wrap a string around the disk. As you pult the end of the string with a steady force, it will unwind and therefore move a longer distance than a string applying the same force but attached to the centre of the disk.

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user mike stone

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Can't it also be argued that, for the given distance $d$ in the diagram, the rotational work is $\tau\theta$ where $\tau$ is the torque and $\theta$ is the angular displacement in radians, whatever that comes out to be?

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Bob D

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It's really subtle! So if I understand correctly, even if the center of mass travels the same distance, the tangential force would travel a greater distance than the force applied to the center of mass? The example of the rope makes it really clear.

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Pierre

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@PierreChamoun Properly speaking a force doesn't travel (this idea can generate confusion in situations like the present one). It's the matter on which it's applied that travels.

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user pglpm

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@pglpm Of course, but don't you think it is a little bit ambiguous to talk about the distance travelled by the particles at the circumference since they are in contact with the force for very little time? But I guess we could use the integral for that: W = ∫F•dx = ∫F•vdt

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Pierre

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@PierreChamoun Indeed, we actually have a force field $\pmb{F}(\pmb{x},t)$. In the reference frame where the disk is at rest, the tangential force appears as a localized, rotating force field. In continuum mechanics, the total rate of work done by the (bulk) forces on a body is given by $\iiint_{\text{body}}\pmb{F}(\pmb{x},t)\cdot\pmb{v}(\pmb{x},t)\ \mathrm{d}\pmb{x}$, similar to what you write

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user pglpm

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@BobD yes, absolutely!

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Dale

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@PierreChamoun said “But I guess we could use the integral for that: W = ∫F•dx = ∫F•vdt”. Indeed, I consider power $P=F\cdot v$ to be the more valuable concept. There are three conserved quantities of interest: energy, momentum, and angular momentum. Power is the rate of change of energy, force is the rate of change of momentum, and torque is the rate of change of angular momentum.

This post imported from StackExchange Physics at 2025-01-22 11:04 (UTC), posted by SE-user Dale