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I hope you are doing well!

I found the following question in Khan Academy:

Steven applies a force $F$ to a disc, halfway between its axle and outer edge, at $r/2$ where $r$ is the radius of the disc. The disc can rotate without friction around its center. Would rotating the disc at a distance of $r$ from the center increase or decrease the angular acceleration?

The correct answer was that increasing the distance to which Force is applied would increase the angular acceleration.

Since Angular Acceleration = Torque/Moment of Inertia, so $α = τ/I$. Since the angle the force is applied is $90^\circ$, we know $τ = rF$, and since the rotating object is a disc, the Moment of Inertia is $I = 0.25mr^2$. Thus, $α = (rF)/(0.25mr^2) = 4F/mr$. As such, wouldn't increasing $r$, the distance to which force is applied, decrease the angular acceleration since $r$ is in the denominator?

It would be awesome if someone could clear my doubt. Thanks, and have a great day!

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  • $\begingroup$ $r$ is the disk radius and it does not change, but the location of the force needs another variable, say $d=r/2$ and work out the problem exactly as you did, but with the torque $\tau =d \, F$. $\endgroup$
    – John Alexiou
    Commented Dec 8, 2020 at 1:59

3 Answers 3

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You're confusing your r's. Let $I=\dfrac{1}{2} mR^2$ where $R$ is the radius of the disk.

Now, say you apply a force $F$ a distance $r$ away from the axel. Since $\tau = Fr$, the angular acceleration is

$$\alpha =\dfrac{\tau}{I} = \dfrac{Fr}{\tfrac{1}{2}mR^2} \quad\implies\quad \alpha =\dfrac{2Fr}{mR^2} $$

Therefore, increasing $r$ (the distance between axel and application point) will increase $\alpha$.

If you change $R$, then what you're actually doing is changing the size of your disk.

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  • $\begingroup$ Oh ok, thanks so much! So if we have an object attached to a string, would the Moment of Inertia of the object depend on the distance it is away from the point where the string rotates, instead of the entire length of the string? $\endgroup$
    – Yashas Ravi
    Commented Dec 7, 2020 at 21:22
  • $\begingroup$ Please be more specific. $\endgroup$
    – user256872
    Commented Dec 7, 2020 at 21:28
  • $\begingroup$ So if there is a string of length L and point mass A is attached to the string at a distance of L, and point mass B attached at a distance of 0.5L, and the point masses are rotated (where the centripetal force is the tension in the string), would the Moment of Inertia of point A be 4 times the Moment of Inertia of point B? $\endgroup$
    – Yashas Ravi
    Commented Dec 7, 2020 at 21:35
  • $\begingroup$ Moment of inertia of point mass is $I=mR^2$ where $R$ is the distance from the center of rotation to the point mass. The system's moment of inertia would be the sum of the moments of inertia of masses A and B (provided their moments of inertia are calculated about the same point and axis, and in this case, they would be) $\endgroup$
    – user256872
    Commented Dec 7, 2020 at 21:41
  • $\begingroup$ Oh ok. So for objects like discs, hoops, cylinders, and spheres, the Moment of Inertia depends on the radius of the whole rotating object, while for a point mass, the Moment of Inertia depends on the distance from the point mass and where the centripetal force is directed? $\endgroup$
    – Yashas Ravi
    Commented Dec 7, 2020 at 21:43
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The $r$ in your formula for the moment of inertia is the radius of the disk. It doesn't change just because you move the point where you're applying force to the disk.

The $r$ in your formula for applied torque is the distance from the axis where the force is applied.

You shouldn't be using the same symbol for these two things in the same problem, and you can't cancel one with the other when you combine the two formulas to get the angular acceleration.

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Yashas Ravi
    Commented Dec 7, 2020 at 21:22
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This is a good time to start thinking about the boundaries of the problem. We can stipulate that if you apply the force at $R$, you get an acceleration $\alpha$. Now if you go to the other side of the disk ($-R$), symmetry tells you're going to get $-\alpha$.

Also: if you apply the force at $R=0$, then nothing spins, $\alpha=0$.

Now plot that in your head (or on paper):

enter image description here

Here I've shown the linear, cubic, and quintic interpolations (even powers are ruled out by symmetry). It's very hard to imagine cubic or quintic to be right, but either way, the magnitude at $x=R/2$ is smaller than at $R$ for any sensible interpolation.

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