---
title: "Sphere strikes hinged rod, e = 0.890"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:sphere-strikes-hinged-rod-2-5kg"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/sphere-strikes-hinged-rod-2-5kg
site: BCT Engineering
---

# Sphere strikes hinged rod, e = 0.890

Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods

- Asked in: 2075 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/sphere-strikes-hinged-rod-2-5kg
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

A 2.5-kg sphere moving horizontally to the right with an initial velocity of 7 m/s strikes the lower end of a 10-kg rod AB. The rod is suspended from a hinge at A and is initially at rest. Knowing that the co-efficient of restitution between the rod and the sphere is 0.890, determine the angular velocity of the rod and the velocity of the sphere immediately after the impact. [Figure: rod AB of length 1.2 m hanging from hinge A; sphere approaching the lower end B horizontally (figure labels the speed 5 m/s).]

## Short answer

Conserve angular momentum about hinge A and use the restitution relation; neglect impulsive hinge reactions and weights. With I_A=1/3(10)(1.2)^2, the rod angular velocity is 4.73 rad/s and the sphere velocity after impact is 0.56 m/s to the left (v'=-0.56 m/s).

## Answer

Take moments about the hinge $A$: during the impact the hinge reaction passes through $A$ and the weights are non-impulsive, so the angular momentum about $A$ of the sphere and rod together is conserved. Let $L = 1.2$ m be the distance from $A$ to the point of impact (the lower end $B$), $v_s'$ the sphere's velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

(The figure label shows 5 m/s but the text gives 7 m/s; 7 m/s is used.)

#### Data
$m_s = 2.5$ kg, $v_s = 7$ m/s, $m_r = 10$ kg, $e = 0.89$

$$
I_A = \frac13 m_rL^2 = \frac13(10)(1.2)^2 = 4.800\ \text{kg m}^2
$$

#### Conservation of angular momentum about A
$$
m_sv_sL = I_A\omega + m_sv_s'L \quad (1)
$$

#### Restitution
The velocity of the rod's lower end is $\omega L$:

$$
e = \frac{\omega L - v_s'}{v_s} \;\Rightarrow\; v_s' = \omega L - e\,v_s \quad (2)
$$

#### Solving
Substituting (2) in (1):

$$
m_sv_sL(1+e) = (I_A + m_sL^2)\,\omega
$$

$$
\omega = \frac{2.5(7)(1.2)(1+0.89)}{4.800 + 2.5(1.2)^2} = 4.73\ \text{rad/s}
$$

$$
v_s' = \omega L - e\,v_s = 4.73(1.2) - 0.89(7) = -0.56\ \text{m/s}
$$

**Answer: $\omega = 4.73$ rad/s; just after impact the sphere moves at $0.56$ m/s to the left (its direction is reversed, $v_s' = -0.56$ m/s).**

## Check yourself

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

1. Which quantity is conserved across the impact, and about which point?
   A. Linear momentum of the rod
   B. Kinetic energy of the system
   C. Angular momentum of sphere + rod about hinge A
   D. Angular momentum about the sphere's centre
   Answer: C. The hinge reaction passes through A and the weights are non-impulsive, so angular momentum about A is conserved; kinetic energy is lost since e < 1. (Key point: Conservation law used)

2. What is the rod's moment of inertia about A (m = 10 kg, L = 1.2 m)?
   A. (1/12) m L^2 = 1.2 kg m^2
   B. m L^2 = 14.4 kg m^2
   C. (1/3) m L^2 = 4.8 kg m^2
   D. (1/2) m L^2 = 7.2 kg m^2
   Answer: C. A slender rod about its end has I = (1/3) m L^2 = (1/3)(10)(1.44) = 4.8 kg m^2. (Key point: Moment of inertia)

3. How is the sphere's final velocity related to w through restitution?
   A. v_s' = w L - e v_s
   B. v_s' = w L + e v_s
   C. v_s' = e v_s - w
   D. v_s' = w / L
   Answer: A. The relative speed of separation w L - v_s' equals e v_s, so v_s' = w L - e v_s. (Key point: Restitution at the rod end)

4. What do the results w = 4.73 rad/s and v_s' = -0.56 m/s mean?
   A. The sphere continues forward at 0.56 m/s
   B. The rod rotates at 4.73 rad/s and the sphere rebounds at 0.56 m/s
   C. The sphere stops and the rod rotates
   D. The rod rotates at 0.56 rad/s
   Answer: B. The negative sign means the sphere's velocity is reversed, since the rod end moves faster (5.67 m/s) than the sphere would continue. (Key point: Final result and meaning)

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