3
$\begingroup$

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.

enter image description here

$\endgroup$
1
2
$\begingroup$

From what I understand, in one case you imagine a force with constant direction, orientation, and magnitude to be tangentially applied, at each instant, on the part of the disk that has highest vertical coordinate with respect to an observer in a reference frame in which the disk is initially at rest. This means that such a force (field) is applied to a different part of the disk at each instant.

Let's fix such a reference frame. In both situations, the balance of linear momentum says that the motion of the centre of mass will be the same. So $\pmb{v}_{\text{cm}}(t)$ is the same in either case.

In the case of the application on the centre of mass, all parts of the disk have the same motion. In particular, the spot on the upper part of the disk has $\pmb{v}_{\text{upper}}(t)=\pmb{v}_{\text{cm}}(t)$ at all $t$.

But in the case of the tangential application, there will also be a torque, and therefore a rotation of the disk. In this case the spot instantaneously in the upper part the disk has higher velocity than the centre of mass (both parallel): $\lvert\pmb{v}_{\text{upper}}(t)\rvert>\lvert\pmb{v}_{\text{cm}}(t)\rvert$.

The instantaneous rate of work (power) done by a force on a small element of matter at time $t$, in a specific reference frame, is given by the scalar product of the force and the instantaneous velocity of that small element: $\pmb{F}(t)\cdot\pmb{v}_{\text{element}}(t)$.

The tangential force therefore does more work, because it's applied to an element (the part of the disk border that's instantaneously in the upper position) that has higher velocity than the element located at the centre of mass: $\pmb{F}\cdot\pmb{v}_{\text{upper}}(t) >\pmb{F}\cdot\pmb{v}_{\text{cm}}(t)$ at each instant $t$.

This is why in the tangential-application case the body has additional rotational energy after the centre of mass has been displaced by the same amount.

This problem is probably clearer when seen from the point of view of continuum mechanics, so I'll give this reference:

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
12
  • 2
    $\begingroup$ 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. $\endgroup$ – mike stone Dec 20 '20 at 19:14
  • 1
    $\begingroup$ 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? $\endgroup$ – Bob D Dec 20 '20 at 19:55
  • $\begingroup$ 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. $\endgroup$ – Pierre Dec 20 '20 at 20:13
  • $\begingroup$ @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. $\endgroup$ – pglpm Dec 20 '20 at 20:33
  • 1
    $\begingroup$ @Dale Just want to make sure I understand your last paragraph. Why do you say the tangential force will be "applied only briefly"? And are you saying that if the center of mass of both disks travel the same distance 𝑑, then the total energy of both disks will be the same, except that the energy of the tangential disk is split between the translational KE and rotational KE? $\endgroup$ – Bob D Dec 21 '20 at 17:13
0
$\begingroup$

One way to approximate what happens to our intuition is to use the equivalence principle of relativity, and let gravity plays the role of the constant force in the space.

The case of a tangential force is like a yo-yo toy, that after being released, the string is carefully pushed upwards to keep the disk at the same height. As its acceleration is zero, the force in the string must balance its weight.

The case of centered force corresponds to the string of the toy totally unwinded, and just kept by the string in the same height. The force is the same as before.

For the first case, there is work because the force is moving: we raise our hand with the string to keep the disk in the same position. All the work is translated in rotational kinetic energy.

Now, from the point of view of an observer in free fall, besides the rotational kinetic energy of the first case, there is also translational kinetic energy for both. But for the same time interval, he measures a greater displacement (of the force, not of the toy) in the first case. That extra displacement corresponds to the additional work to rotate the toy.

By the equivalence principle, that is exactly the situation to an inertial observer in the outer space, as stated by OP.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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