303 - Mechanics part 1 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.

$$\begin{gathered} According to given condition 2r = {u_y}^2/2g, \\ T= 2u_y/g =2\sqrt{4gr}/g=4\sqrt{r/g} \\ At top point due to touch, vecocity of the frog be u_x-2v. \\ R=[u_x{u}_y/g]+[u_x-2v)u_y/g]=[u_y/g\left]\right(2u_x-2v) \\ =[2u_y/g\left]\right(u_x-v) =T(u_x-v) \\ Now minimum possible value of velocity of frog at top to \\ cross the cylinder is \sqrt{gr} (By the concept of vertical circle) \\ now solve carefully without error \end{gathered}$$

$$\begin{gathered} Let the acceleration of hemisphere is a_h \\ N \sin \theta = M{a}_h \\ \Rightarrow N/m{a}_h = M/m\sin \theta \\ Here pseudo force is considered due to acceleration of hemisphere \end{gathered}$$

$$\begin{gathered} In first case F = 2T = 150 \\ so T = 75 newton, \\ Now in second case system is free to move \\ acceleration of the system a = F/20 \\ for 10 kg rod F-2T = 10a = 10(F/20) = F/2 \\ \Rightarrow F-(F/2)=2T \Rightarrow F=4T = 300 newton \end{gathered}$$

$$\begin{gathered} Moment of inertia of a cube about a line perpendicular to \\ its one face and pas\sin g through its centre is m{a}^2/6 \\ so about diagonal inertia = l^2{I}_x+m^2{I}_y+n^2{I}_z \\ = (l^2+m^2+n^2)I_x (here I_x=I_y=I_z) \\ = I_x = m{a}^2/6 \\ l,m,n are direction \cos ines of three lines pas\sin g througe \\ centre of cube and perpendicular to each other \end{gathered}$$

Problem is interesting yet not complete. Complete it to get answer 

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

$$\begin{gathered} let u and v are the speeds of boy and escalator \\ according to given condition \\ \left[1\right(u+v)T-2(u-v\left)T\right]n_1=\left[2\right(u+v)T-1(u-v\left)T\right]n_2 \\ \Rightarrow \left[1\right(u+v)-2(u-v\left)\right]T{n}_1=\left[2\right(u+v)-1(u-v\left)\right]T{n}_2 \\ Given that T{n}_1=250 and T{n}_2=50 \\ \Rightarrow \left[1\right(u+v)-2(u-v\left)\right]250=\left[2\right(u+v)-1(u-v\left)\right]50 \\ now solve it for v, u is given \end{gathered}$$