# Use of Instantaneous axis of rotation (IAOR) in a question of mechanics give different and wrong answer

The above situation can be solved easily when we go in frame of point C then point C will come at rest and mass "m" will do circular motion about point C with 3v velocity and then using Newton's 2nd law and concept of centripetal acceleration

T - mg = m(3v)²/l

T = mg + 9mv²/l

but if we use the concept Instantaneous axis of rotation -IAOR which is at a distance of 2L/3 from mass m for this situation then tension will come out to be mg + m(2v)²/(2L/3) or mg + 6mv²/l. I am not getting that where I am lacking in my concepts. Please clear my misconceptions this is not a homework or check my work question.

• This question actually asks about a conceptual point, not about how to calculate the answer. As such, it really should be permissible. Jan 17, 2023 at 17:35
• I voted to reopen after the edit, which brough out the conceptual core of the question quite clearly. (Sorry if I was a bit too overzealous regarding the first version). Jan 17, 2023 at 20:23

This does not work out, because the radius in centripetal acceleration formula is determined by the radius of curvature of the path of the mass point in an inertial frame.

However, the distance to the IAOR does not give the right radius of curvature because the motion of the points $$C$$ and $$D$$ around the considered point in time is only approximated by a rigid body rotation around the IAOR to the first order, but not to the second order (which determines the centripetal force).

This can be easily seen as follows: If they were moving as if rotating around the IAOR then the $$y$$-coordinate of $$C$$ would have to decrease (which it doesn't). In turn the $$y$$-coordinate of $$D$$ increases to little to keep the string length constant when assuming rotation around the IAOR to second order.

More explicitly, this can be seen by looking at the configuration of the system a small time increment after the situation in both cases.

If you use the frame where $$C$$ does not move, after a small time increment $$dt$$, the mass point's $$x$$-coordinate will be $$x_D(dt) = x_C(0) + 3v_0 dt$$ (since $$dt$$ is small any higher order effects due to the change of velocity during the time interval can be neglected). Now the string gives you the constraint, that at all times $$l = \sqrt{(x_D - x_C)^2 + (y_D - y_C)^2}$$, so we can derive $$y_D(dt) = \sqrt{l^2 - x_D^2} = l \sqrt{ 1 - \frac{x_D^2}{l^2}} \approx l - \frac{1}{2} \frac{9v_0^2}{l} dt^2.$$ From this, we can directly read off the centripetal acceleration: $$a = \frac{9v_0^2}{l}.$$

If we do the same analysis in the frame with the IAOR we find the following: $$x_D(dt) = x_C(0) + 2 v_0 dt$$ $$x_C(dt) = x_C(0) - v_0 dt$$ $$y_C(dt) = y_C(0)$$ $$y_D(dt) = \sqrt{l^2 - (x_D(dt) - x_C(dt))} = l \sqrt{1 - \left(\frac{3v_0dt}{l} \right)^2 }$$ So the string length constraint, plus the same relative speed between $$C$$ and $$D$$ leads to the same result for $$y_D(t)$$ and in turn the same centripetal acceleration in both frames.

Note that there is a principle behind the fact that the system where $$C$$ is fixed gives an easy, geometric way to determine the radius of curvature of the paths of the masses.

Namely, that it is the center of mass system (since the described system behaves like to point masses connected by a string, as the mass $$m_C \to \infty$$).