Question

A uniform rod of mass $M$ and length $2L$ lies on a smooth horizontal table. There is a smooth peg $O$ is fixed on the table. One end of the rod is kept touching the peg as shown in the figure. An impulse $J$ is imparted to the rod at its other end. The impulse is horizontal and perpendicular to the length of the rod. The magnitude of impulse experienced by the peg, if $J = 80$ N-sec is

• A. 10 N-sec
• B. 20 N-sec
• C. 30 N-sec
• D. 40 N-sec

Answer 1

Answer: (D)

40 N-sec

Answer 2

Let the rod have mass $M$ and length $2L$. The rod is pivoted at one end $O$. An impulse $J$ is applied at the other end, perpendicular to the rod. Let the impulse from the peg on the rod be $J_{peg}$.

We apply the angular impulse-angular momentum theorem about the pivot point $O$. The moment of inertia of a uniform rod of mass $M$ and length $2L$ about one end is $I_O = \frac{1}{3}M(2L)^2 = \frac{4}{3}ML^2$. The impulse $J$ is applied at a distance $2L$ from the pivot. The angular impulse due to $J$ about $O$ is $(2L)J$. The impulse $J_{peg}$ is applied at the pivot $O$, so its lever arm is zero, and it contributes no angular impulse about $O$. The initial angular momentum about $O$ is zero. Let $\omega$ be the angular velocity of the rod after the impulse. The final angular momentum about $O$ is $I_O \omega$. Thus, by the angular impulse-angular momentum theorem: $(2L)J = I_O \omega$ $(2L)J = \frac{4}{3}ML^2 \omega$ $\omega = \frac{2LJ}{\frac{4}{3}ML^2} = \frac{3J}{2ML}$

Now, we apply the linear impulse-momentum theorem to the center of mass (CM) of the rod. The CM of the rod is at a distance $L$ from the pivot $O$. The velocity of the CM is $v_{CM} = \omega \times L = \omega L$. $v_{CM} = \left(\frac{3J}{2ML}\right) L = \frac{3J}{2M}$

The total impulse acting on the rod is the sum of the applied impulse $J$ and the impulse from the peg $J_{peg}$. This total impulse equals the change in linear momentum of the rod. $J + J_{peg} = M v_{CM}$ Substituting the expression for $v_{CM}$: $J + J_{peg} = M \left(\frac{3J}{2M}\right)$ $J + J_{peg} = \frac{3}{2}J$ $J_{peg} = \frac{3}{2}J - J = \frac{1}{2}J$

The magnitude of the impulse experienced by the peg is $|J_{peg}| = \frac{1}{2}J$. Given $J = 80$ N-sec, $|J_{peg}| = \frac{1}{2}(80 \text{ N-sec}) = 40 \text{ N-sec}$.