11 - Centre of mass and momentum conservation Questions Answers

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Initially there is no external force. Centre of mass of the system accelerates when insect moves with acceleration.

Now suppose displacement of insect at t = t is x with respect to rod, and y is the acceleration of rod in downward direction with respect to ground. Thus acceleration of counterweight is y with respect to ground. Now$\overrightarrow{x_{CM}}=\left[m\right(x-y)\hat{j}-(M-m)y\hat{j}+My\hat{j}]/2M$Now solve for x = l,

For dynamics, take friction upward in free body diagram of insect and downward in the free body diagram of rod.

Now solve

Inform for any issue.

$$\begin{gathered} At a general point \\ -2T\sin \theta = ma (\theta is measured from horizontal) \\ \Rightarrow a = -2Mg\theta /m = -2Mgy/ml \\ compare with a = -{\omega }^{2}y \\ \omega =\sqrt{\frac{2Mg}{ml}} \left(1\right) \\ Now for max displacement of small m \\ mg(h+y) = 2Mgx and x = y^2/2l \\ on solving this quadratic eq. for y/l we get \\ \frac{y}{l}-\frac{m}{2M}=\sqrt{\frac{m}{M}\frac{h}{l}}\Rightarrow \frac{y}{l}=\sqrt{\frac{m}{M}\frac{h}{l}} (given that \frac{m}{M} <<\frac{h}{l}) \\ so x = \frac{m}{2M}h \left(2\right) \\ so v_{max} = x\omega \\ now solve \end{gathered}$$

distance covered in n sec. = un+an²/2   (1)

distance covered in n-1 sec. = u(n-1)+a(n-1)²/2    (2)

so s_n = (1) - (2)

put equation (1) and (2) and get the answer 

for an axis parellel to the axis of rod and at distance l, rod will act as a set of equidistance particles. so answer will be Ml²

answer given by sarika is not completely correct

velocity of impact = √2gh

so components of velocity after rebound = √2ghcosθ perpendicular to the plane and √2ghsinθ parallel to the plane for the collision is perfectly elestic

now with perpendicular component take net displacement 0 and calculate time 

then find displacement along the plane by using s = ut + 1/2at²

here remember that you should divide gravity in comnents too