A uniform rod of mass 3 kg and length…

A uniform rod of mass 3 kg and length 1.2 m hangs at rest from a smooth horizontal hinge at its upper end. A lump of putty of mass 1 kg moving…

Numerical answer

A uniform rod of mass

3\ \mathrm{kg}

and length

1.2\ \mathrm{m}

hangs at rest from a smooth horizontal hinge at its upper end. A lump of putty of mass

1\ \mathrm{kg}

moving horizontally at

6\ \mathrm{m\,s^{-1}}

, in the vertical plane of the rod, strikes the lower end of the rod and sticks to it. Take

g=10\ \mathrm{m\,s^{-2}}

. The maximum angle (in degrees) through which the rod with the putty swings from the vertical is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 45.57

The hinge force acts at the pivot, so angular momentum about the hinge is conserved during the collision:

m v L=\left(\tfrac13ML^2+mL^2\right)\omega

. With

L=1.2

I=\tfrac13(3)(1.44)+1.44=2.88\ \mathrm{kg\,m^2}

and

\omega=\dfrac{1\times6\times1.2}{2.88}=2.5\ \mathrm{rad\,s^{-1}}

. After the collision, only gravity does work (the hinge is smooth), so

\tfrac12I\omega^2=\left(M\tfrac L2+mL\right)g(1-\cos\theta)

. The left side is

\tfrac12(2.88)(6.25)=9\ \mathrm{J}

and the bracket is

(3\times0.6+1.2)\times10=30\ \mathrm{J}

, so

1-\cos\theta=0.3

\cos\theta=0.70

and

\theta=\cos^{-1}0.70=45.57^\circ

. Treating linear momentum as conserved would be wrong, since the hinge supplies an impulse.

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Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake