Impulsive Collisions in Rotation | JEE Notes

A collision contains large forces acting for a short time. Replace an unknown force history by impulse $𝐉 = \int 𝐅 d t$ . Linear impulse changes momentum; angular impulse changes angular momentum:

Δ 𝐏 = 𝐉_{e x t}, Δ 𝐋_{O} = \int 𝛕_{O, e x t} d t .

Use a fixed inertial point $O$ , or the system centre of mass with the appropriate centre-of-mass angular momentum. Ordinary gravitational impulse is often negligible over the short collision interval, but a hinge can supply a large impulse. “The collision is brief” does not justify ignoring the hinge.

A projectile embeds in a hinged rod

A uniform rod, initially hanging at rest, has mass $M$ , length $L$ , and a fixed hinge at its upper end $O$ . A projectile of mass $m$ hits perpendicular to the rod at distance $a$ from the hinge, speed $u$ , and sticks. During impact choose rod plus projectile as the system and $O$ as the angular-momentum origin. The hinge impulse has zero moment about $O$ .

Watch the animation: Bullet into a hinged rod

m u a = (I_{O} + m a^{2}) ω, I_{O} = \frac{13}{M} L^{2},

ω = \frac{m u a}{I_{O} + m a^{2}} .

Linear momentum of this system is generally not conserved because the hinge is external. Kinetic energy is also not conserved because the impact is inelastic. After impact, however, an ideal fixed hinge does no work, and mechanical energy can determine the subsequent rise:

\frac{12}{(} I_{O} + m a^{2}) ω^{2} = g (M L / 2 + m a) (1 - \cos θ_{\max}) .

This turning-angle equation applies when the energy is below the energy needed to pass the highest position; check that its implied cosine lies in the physical range.

Try the simulator: Bullet into a hinged rod

Example 23 · Impact first, swing second

A $0.60 k g$ , $1.0 m$ uniform rod hangs from one end. A $0.20 k g$ clay projectile moving horizontally at $10 m s^{- 1}$ embeds at the lower end. Find the immediate angular speed, collision energy loss, and maximum angle from downward vertical. Take $g = 10 m s^{- 2}$ .

Hint

Conserve angular momentum about the hinge during impact. Use energy only after impact.

Complete solution

$I_{t o t a l} = 0.60 (1)^{2} / 3 + 0.20 (1)^{2} = 0.40 k g m^{2}$ . Hence $ω = (0.20) (10) (1) / 0.40 = 5 r a d s^{- 1}$ . Initial projectile energy is $10 J$ ; post-impact rotational energy is $\frac{12}{(} 0.40) (25) = 5 J$ . The collision loses $5 J$ .

During the swing, $5 = [0.60 (10) (0.50) + 0.20 (10) (1)] (1 - \cos θ) = 5 (1 - \cos θ)$ . Therefore $θ_{\max} = 90^{\circ}$ .

Why the tempting shortcut fails: using the projectile's full $10 J$ for the rise silently restores the energy dissipated during embedding.

If the projectile rebounds: restitution concerns contact velocities

For the same initially stationary hinged rod, let $J$ be the impulse transferred in the incident direction and $e$ the normal coefficient of restitution. The projectile emerges with signed velocity $v = u - J / m$ ; the struck rod point moves at $a ω$ , with $I_{O} ω = J a$ . Restitution gives $a ω - v = e u$ , so

J = \frac{(1 + e) u}{1 / m + a^{2} / I_{O}} .

Restitution is not a relation between the projectile and the rod centre. It uses the two material points at contact, immediately before and after impact. For an oblique collision, apply it only along the specified collision normal and model tangential impulse separately.

A free rod can both translate and rotate

Now place the initially stationary rod on a smooth horizontal table, without a hinge. A projectile travels perpendicular to the rod and hits a distance $a$ from its centre. Define positive rod rotation so the struck point gains velocity along the incident direction. With $I_{C} = M L^{2} / 12$ ,

Watch the animation: Centre of percussion

Try the simulator: Hit on a free rod

V = \frac{J}{M}, ω = \frac{J a}{I_{C}}, v = u - \frac{J}{m} .

V + a ω - v = e u, J = \frac{(1 + e) u}{1 / m + 1 / M + a^{2} / I_{C}} .

The denominator is the inverse effective collision mass. The term $a^{2} / I_{C}$ shows why an off-centre struck point can recoil strongly by rotating, even if centre translation is modest. The smooth table supplies no horizontal impulse; linear momentum in the incident direction is conserved. For $a = 0$ , the rotation term disappears and the familiar one-dimensional result returns.

For this initially stationary rod and a normal impact, define $m_{e f f} = [1 / m + 1 / M + a^{2} / I_{C}]^{- 1}$ . The kinetic energy dissipated is ${\frac{12}{m}}_{e f f} (1 - e^{2}) u^{2}$ . It vanishes at $e = 1$ and is largest at $e = 0$ . A zero restitution coefficient means zero immediate relative normal velocity at contact; permanent attachment additionally requires the physical statement that the projectile embeds or sticks.

Example 24 · The struck end moves faster than the incoming particle

A $1 k g$ particle at $5 m s^{- 1}$ strikes one end of a stationary $1 k g$ , $2 m$ uniform rod on a smooth table. Motion is perpendicular to the rod, and the collision is perfectly elastic. Find both outgoing motions.

Hint

Here $a = 1 m$ , $I_{C} = 1 / 3 k g m^{2}$ , and the contact-point speed is $V + a ω$ .

Complete solution

$J = 2 (5) / (1 + 1 + 3) = 2 N s$ . Thus rod-centre speed $V = 2 m s^{- 1}$ , rod angular speed $ω = 2 (1) / (1 / 3) = 6 r a d s^{- 1}$ , and particle speed $v = 5 - 2 = 3 m s^{- 1}$ , still forward.

The struck end has instantaneous speed $V + a ω = 8 m s^{- 1}$ . Its separation speed relative to the particle is $8 - 3 = 5 m s^{- 1}$ , equal to the approach speed. Energy checks: $K_{f} = \frac{12}{(} 1) (3^{2}) + \frac{12}{(} 1) (2^{2}) + \frac{12}{(} 1 / 3) (6^{2}) = 12.5 J = K_{i}$ .

What is conserved?

Situation	$\vec{p}$	$L$	$K$
Particle sticks to hinged rod	No	About hinge	No
Particle sticks to free rod	Yes	About any fixed point	No
Elastic hit on free rod	Yes	About any fixed point	Yes
Elastic hit on hinged rod	No	About hinge	Yes

Centre of percussion

A hinged rod struck at distance $a$ from the hinge gets hinge impulse $H = J (\frac{3 a}{2 L} - 1)$ . It vanishes at $a = \frac{23}{L}$ . In general $a = \frac{I_{O}}{M d}$ . This is the sweet spot of a cricket bat.

Rotational collisions: choose the system and the axis first

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