---
title: "Ball striking a pivoted rod"
subject: "Engineering Mechanics"
chapter: 8
key: "semester_3/engineering_mechanics:ball-strikes-hanging-rod-eccentric"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/ball-strikes-hanging-rod-eccentric
site: BCT Engineering
---

# Ball striking a pivoted rod

Engineering Mechanics · Chapter 8: Plane Motion of Rigid Bodies: Impulse and Momentum Method

- Asked in: year not given, 8 marks
- Video: [Angular Velocity and Displacement of a Rod Struck by a Ball](https://www.youtube.com/watch?v=eCdNB5F7sw0) — INTEGRAL PHYSICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/ball-strikes-hanging-rod-eccentric
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-8

## Question

A uniform slender rod of mass 3 kg and length 1.2 m hangs vertically at rest from a smooth horizontal pin at its upper end O. A ball of mass 0.5 kg moving horizontally at 10 m/s strikes the lower end of the rod. The coefficient of restitution is 0.6. Find the angular velocity of the rod and the velocity of the ball just after impact, and the kinetic energy lost.

## Short answer

Apply conservation of angular momentum about the pivot and coefficient of restitution. With I_O = ML²/3 = 1.44 kg·m², solving gives angular velocity 4.44 rad/s, ball rebounds at 0.67 m/s, and kinetic energy lost 10.67 J.

## Answer

**Given:** rod $M = 3$ kg, $L = 1.2$ m, pivot O (smooth); ball $m = 0.5$ kg, $u = 10$ m/s, strikes at the free end (distance $L$ from O); $e = 0.6$.

```text
     O (pin)
     |
     |  rod M
     |
     |
  -->o  ball m, u = 10 m/s     (impact at the lower end)
```

#### Moment of inertia of the rod about O

$$
I_O = \frac{ML^2}{3} = \frac{3(1.2)^2}{3} = 1.44\ \text{kg·m}^2
$$

#### Equations
Let $v$ be the velocity of the ball after impact (positive in the original direction) and $\omega$ the angular velocity of the rod after impact. The pin reaction is impulsive, but it passes through O, so **angular momentum about O is conserved** for the system of ball plus rod:

$$
m u L = m v L + I_O\omega
$$

$$
0.5(10)(1.2) = 0.5\,v\,(1.2) + 1.44\,\omega \;\Rightarrow\; 6 = 0.6v + 1.44\omega \quad (1)
$$

Restitution (relative speed of separation at the contact point, rod tip speed $= \omega L$):

$$
\omega L - v = e\,u = 0.6(10) = 6 \;\Rightarrow\; v = 1.2\omega - 6 \quad (2)
$$

Substituting (2) in (1): $6 = 0.6(1.2\omega - 6) + 1.44\omega = 0.72\omega - 3.6 + 1.44\omega$

$$
9.6 = 2.16\,\omega \;\Rightarrow\; \omega = 4.444\ \text{rad/s}
$$

$$
v = 1.2(4.444) - 6 = -0.667\ \text{m/s}
$$

The negative sign means the ball rebounds (moves back) at 0.667 m/s. The rod tip moves at $\omega L = 5.333$ m/s.

#### Kinetic energy lost

$$
T_1 = \tfrac12(0.5)(10)^2 = 25.00\ \text{J}
\qquad
T_2 = \tfrac12(0.5)(0.667)^2 + \tfrac12(1.44)(4.444)^2 = 14.33\ \text{J}
$$

$$
\Delta T = 10.67\ \text{J}
$$

**Answer: ω = 4.44 rad/s; ball rebounds at 0.67 m/s; energy lost = 10.67 J.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is angular momentum about O conserved during impact?
   A. Energy is conserved
   B. The pin reaction is impulsive but passes through O, so it has no moment about O
   C. The pin is smooth so there is no reaction
   D. The ball is light
   Answer: B. No external angular impulse about O acts on ball plus rod. (Key point: Conserved quantity at the pin)

2. Which equation expresses conservation of angular momentum about O?
   A. m u L = m v L + I_O omega
   B. m u = m v + M omega L
   C. m u L = I_O omega only
   D. m u = m v
   Answer: A. 6 = 0.6 v + 1.44 omega. (Key point: Moment equation)

3. Which is the restitution relation at the contact point?
   A. omega L - v = e u
   B. omega L + v = e u
   C. omega L = u
   D. v = e u
   Answer: A. Separation speed of rod tip and ball equals e times approach speed, so omega L - v = 6. (Key point: Restitution equation)

4. What happens to the ball after impact?
   A. It continues forward at 0.667 m/s
   B. It rebounds at 0.667 m/s, with omega = 4.44 rad/s
   C. It stops
   D. It rebounds at 6 m/s
   Answer: B. Solving gives omega = 4.444 rad/s and v = -0.667 m/s, a rebound. (Key point: Result)

## Related in Engineering Mechanics

- [Rod released from horizontal position](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-release-from-horizontal-reaction) — Engineering Mechanics, chapter 6
- [Oblique impact of ball on a wall](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aball-oblique-impact-wall) — Engineering Mechanics, chapter 4
- [Direct central impact and restitution](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Adirect-central-impact-derivation) — Engineering Mechanics, chapter 4
- [Mass moment of inertia derivations](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Amass-moment-inertia-parallel-axis) — Engineering Mechanics, chapter 6
- [Two blocks connected over a pulley](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ablocks-pulley-friction-acceleration) — Engineering Mechanics, chapter 2
- [Composite body moment of inertia](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-sphere-composite-inertia) — Engineering Mechanics, chapter 6

## Related in other subjects

- [Ball strikes suspended rod, constrained motion](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aball-strikes-suspended-rod-50n) — Applied Mechanics (Dynamics), chapter 6
- [Sphere strikes hinged rod, e = 0.890](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Asphere-strikes-hinged-rod-2-5kg) — Applied Mechanics (Dynamics), chapter 6
- [Sphere strikes hinged rod, e = 0.8](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Asphere-strikes-hinged-rod-3kg) — Applied Mechanics (Dynamics), chapter 6
- [Rod pivoted at O, spring compressed 25 mm](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-spring-pivot-15kg-1-5m) — Applied Mechanics (Dynamics), chapter 6