Question

My questions is concerning the Rotation of Rigid Bodies. Does the choice of Instantaneous Rotation axis changes the answer(angular acceleration and all). For instance how about a Yo-Yo under the force of gravity. In my opinion it shouldn't change the answer but I have book in which author has a fixed way of choosing the Instantaneous Rotation axis.

Answer 1

What would you do in a Problem like this?

Objective is to Determine the acceleration of the Discs and friction is just enough for pure rolling.



Mass is same and so is the radius for both of the discs.

Answer 2

Pulley is mass-less and the string connecting the Discs is in-extensible and light i.e mass-less.

Answer 3

The choice is not fixed, but there are certainly wiser choices than others. In the problem above, are those strings attached to the midpoints of those discs somehow, or are they attached to the edges?

Answer 4

Thanks Can you tell me little about wise-choices, I usually prefer Axis passing through Center of Mass.

Edges...

Answer 5

Hmm I was thinking as Tension would be same for both of the discs, The angular acceleration should be same too.
Axis passing through Center of Mass.
\(I\) for moment of Inertia
\(T\) is for Tension
\[ \sum \tau = rT = I \alpha\]
And I even assumed their linear acceleration to be same (I think this is where I committed mistake), but I am not getting the right answer :/

Answer 6

It varies from problem to problem, but I'd say in general that's not a bad idea. A good choice is one that makes calculation of torques the easiest, often by eliminating some because any force acting on the axis of rotation cannot produce a torque. What answer are you finding?

Answer 7

Hmm I am getting \[a = \frac{1}{2}g \sin{\theta}\]
But I am supposed to get \[\frac{1}{3}g\sin{\theta}\]

Answer 8

For the disc down the slope,
\[mg\sin\theta - T = ma\]For the Disc on the Plane,
\[ma = T\]

Answer 9

Give me a moment to type a full response: