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The question asks what the angular acceleration of an uniform disc of radius $R$ rotating about an axis passing through its edge if it is released from rest with its center of mass at the same height as the axis.

Like this, with the red dot being the axis and yellow dot being center of mass:

Diagram

The angular acceleration is $2g/3R$, because the torque is $MgR$ (cross product of gravitational force times radius vector) and you can divide the torque by the moment of inertia $3/2(MR^2)$ to get the angular acceleration.

However, I'm having trouble seeing why the answer also can't be just $g/R$. At the exact moment that it is released from rest, all parts of the disc (center of mass included) should just be accelerating at $9.81 \text{ m/s}^2$ downwards as the gravitational force is the only force acting upon the disc; hence, why can't you just divide the net tangential acceleration by the radius to get the net angular acceleration as net angular acceleration=net tangential acceleration/R?

Why aren't these two answers equivalent? What am I missing?

Thanks.

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    $\begingroup$ Is the pivot point (red dot) supplying any force to the disc? $\endgroup$ – BowlOfRed Nov 5 '15 at 16:52
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The issue here is that there is a force acting on the axle to counter act gravity on that part of the disk, so not all parts of the disk will be accelerating uniformly at $9.81 \text{ m/s}^2$. Note also that this force acts at the axis of rotation, and therefore doesn't contribute to the torque. That is why you don't need to consider it when calculating the angular acceleration at the center of the disk.

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  • $\begingroup$ Ah, okay. So what would the magnitude of this force be at the axis and at the center of mass? Is there a way to mathematically calculate this force without using equations for angular motion? $\endgroup$ – John Smith Nov 5 '15 at 17:23
  • $\begingroup$ @JohnSmith You can draw a free body diagram, where gravity acts on the center of mass and the force from the axle acts at the point of the axle. I'm not sure that you can calculate the magnitude of the force on the pivot point without using angular motion. The total force on the object should sum to produce the mass times the linear acceleration of the center of mass at that time. You can use the angular acceleration to find the linear acceleration of the center of mass. $\endgroup$ – tmwilson26 Nov 5 '15 at 17:41
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If you have found that the initial angular acceleration of the disc is $\alpha = \frac {2g}{3R}$ that must mean that the initial centre of mass acceleration $a = \frac {2g}{3}$ because $a = R \alpha$.

If the force acting on the pivot is $X$ upwards then applying Newton's second law for the centre of mass motion gives $mg-X = ma$, so $X = \frac {mg}{3}$.

The fact that as well as its weight there is another force acting on the disc at the pivot end and the fact that one end will not accelerate at all whilst the other end will have a linear acceleration twice that of the centre of mass should show you that the initial linear acceleration of the disc is not $g$.

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