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I do not agree with you. In the special case ##V(t)=0## for all ##t##, your expression predicts that the cylinder will be moving. Does that make sense? Besides, what is ##\omega##? Your final answer must be in terms of the given quantities and these are ##M,~k,~R## and ##V(t)##.JD_PM said:To Recap: I am still convinced that the right answer is(see #9 for more details):
$$v(t) = v_{CM} + V(t) = 2R \omega + V(t)$$
Do you agree with me?
The strategy for finding ##v_{cm}(t)## is to use dynamics to find ##a_{cm}(t)## and integrate. ##\vec F_{net}= M\vec a_{cm}## is insufficient to complete the task because there is also angular acceleration. So you need ##\tau=I\alpha##. I am arguing that if you transform to the non-inertial frame, it is sufficient to use ##\tau=I\alpha## in order to find the acceleration of the cm in the non-inertial frame and hence in the inertial frame. However, for that to happen, you need to choose wisely the axis about which you the torque and the moment of inertia are to be expressed.JD_PM said:My solution is obtained from an inertial reference frame.
@kuruman you were interested in solving the problem from a non inertial reference frame. why?
It's not "if we use", it's "because we have to use".JD_PM said:I would say that the axis plays a role if we use:
$$\tau = I \alpha$$
The angular acceleration is obtained from the stated equation:
$$\alpha = \frac{\tau}{I}$$