Question

In the shown figure, a particle of mass 'm' slides down the frictionless surface from height 'h' and collides with the uniform vertical rod of length L and mass M. After the collision, mass m sticks to the rod. The rod is free to rotate in a vertical plane about a fixed axis through O. Find the cosine of the maximum angular deflection of the rod from its initial position.

• A. $1-\frac{m^2{v}_0^2}{2(m+\frac{M}{3})(m+\frac{M}{2})gL}$
• B. $1-\frac{M^2{v}_0^2}{2(m+\frac{M}{3})(m+\frac{M}{2})gL}$
• C. $1-\frac{m^2{v}_0^2}{(m+\frac{M}{3})(m+\frac{M}{2})gL}$
• D. $1-\frac{M^2{v}_0^2}{(m+\frac{M}{3})(m+\frac{M}{2})gL}$

Answer 1

The correct option is A

$1-\frac{m^2{v}_0^2}{2(m+\frac{M}{3})(m+\frac{M}{2})gL}$

Just before collision, velocity of the mass m is along the horizontal and is equal to $v_0=\sqrt{2}gh$. From the FBD during the collision, we note that net torque of impulses about O is zero

∴The angular momentum of the system aboutO is conserved.

If $L_{1}and{L}_2$ are the angular moment a of the system just before and just after the collision, then

$L_1=m{v}_{)}L$

and $L_2=I\omega =(\frac{M{L}^2}{3}+m{L}^2)\omega$

From the conservation of angular momentum,$(\frac{M}{3}+m)L^2\omega =m{v}_{0}L\Rightarrow \omega =\frac{m{v}_0}{(\frac{M}{3}+m)L}$ ...(i)

Let the rod deflects through an angle $\theta$.

initial energy of rod and mass system$=\frac{I}{{\omega }^2},whereI=(\frac{M{L}^2}{3}+m{L}^2)$

Gain in potential energy of the

system=mgl$[1-cos\theta ]+Mg\frac{L}{2}(1-cos\theta )=(m+\frac{M}{2})gL(1-cos\theta )$

From the conservation of energy $=\frac{I}{{\omega }^2}=(m+\frac{M}{2})gL(1-cos\theta )$

putting $\omega$ from (i) in (ii),we get $cos\theta =1-\frac{1}{2}\frac{m^2{v}_0^2}{[\frac{M}{3}+m][\frac{M}{2}+m]gL}$