---
title: "Sphere strikes hinged rod, e = 0.8"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:sphere-strikes-hinged-rod-3kg"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/sphere-strikes-hinged-rod-3kg
site: BCT Engineering
---

# Sphere strikes hinged rod, e = 0.8

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

- Asked in: 2080 Baishakh, 2+8 marks
- Video: [Collision of ball with free rod/ conservation of angular momentum/rotational dynamics](https://www.youtube.com/watch?v=kIUggonZzQI) — Ashwini Jha
- 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-3kg
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

A 3-kg sphere with an initial velocity of 5 m/s strikes the lower end of an 8-kg rod AB. The rod is hinged at A and initially at rest. The coefficient of restitution between the rod and sphere is 0.8. Determine the angular velocity of the rod and the velocity of the sphere immediately after impact. Explain D'Alembert's Principle. [Figure: rod AB hanging from hinge A, centre G at 0.6 m, total length 1.2 m; sphere approaching the lower end B horizontally with $v_s$.]

## Short answer

D'Alembert's principle converts dynamics to statics by adding an inertia force -ma at G and an inertia couple -Iα, giving dynamic equilibrium. Conserving angular momentum about hinge A and using restitution with I_A=1/3(8)(1.2)^2 gives angular velocity 3.97 rad/s and sphere velocity 0.76 m/s in the original direction.

## Answer

**D'Alembert's principle** converts a dynamics problem into a statics problem. A body in motion is in dynamic equilibrium if, in addition to the real forces and moments, we add *inertia effects* equal and opposite to the effective forces:

- Inertia force $-m\bar{\vec a}$ acting at the mass centre $G$.
- Inertia couple $-\bar I\vec\alpha$ (for plane motion).

For a rigid body in plane motion the equations of dynamic equilibrium are

$$
\sum F_x - m\bar a_x = 0, \qquad \sum F_y - m\bar a_y = 0, \qquad \sum M_G - \bar I\alpha = 0
$$

which are the same as $\sum \vec F = m\bar{\vec a}$ and $\sum M_G = \bar I\alpha$, but written as "sum of forces including inertia terms is zero". The moment equation can then be taken about any point, as in statics. Special cases: pure translation ($\alpha=0$): only the inertia force; rotation about a fixed axis through $G$: only the inertia couple $-I\alpha$.

```text
   real forces + (-m a) at G + (-I alpha) couple  =>  equilibrium
```

#### Impact problem
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 sphere + rod 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 velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

#### Data
$m_s = 3$ kg, $v_s = 5$ m/s, $m_r = 8$ kg, $e = 0.8$

$$
I_A = \frac13 m_rL^2 = \frac13(8)(1.2)^2 = 3.840\ \text{kg m}^2
$$

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

#### Restitution
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):

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

$$
\omega = \frac{3(5)(1.2)(1+0.8)}{3.840 + 3(1.2)^2} = 3.97\ \text{rad/s}
$$

$$
v_s' = \omega L - e\,v_s = 3.97(1.2) - 0.8(5) = 0.76\ \text{m/s}
$$

**Answer: $\omega = 3.97$ rad/s; the sphere's velocity just after impact is $0.76$ m/s (in the original direction).**

## Check yourself

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

1. What does D'Alembert's principle do?
   A. It makes the net force on a body zero in all cases
   B. It converts a dynamics problem into a statics problem by adding inertia force and inertia couple
   C. It removes the need for any force
   D. It applies only to particles at rest
   Answer: B. Adding the inertia force -m a at G and the inertia couple -I alpha to the real forces gives dynamic equilibrium, so statics equations can be used. (Key point: D'Alembert's principle)

2. For a rigid body in plane motion, what inertia effects are added?
   A. Only a force at G
   B. Only a couple
   C. A force m a at G in the same direction as a and a couple I alpha in the same sense as alpha
   D. A force -m a at G and a couple -I alpha
   Answer: D. The inertia force is -m a at G and the inertia couple is -I alpha, opposite to the effective force and effective couple. (Key point: Inertia effects)

3. Why is angular momentum about hinge A conserved during the impact?
   A. The hinge reaction acts through A and the weights are non-impulsive
   B. Linear momentum is conserved for the rod
   C. The collision is perfectly plastic
   D. The sphere does not rotate
   Answer: A. No impulsive moment acts about A, so angular momentum of sphere + rod about A is the same before and after. (Key point: Conserved quantity in impact)

4. Combining momentum and restitution, which expression gives the rod's angular velocity?
   A. w = m_s v_s L (1+e) / (I_A + m_s L^2)
   B. w = m_s v_s L / I_A
   C. w = m_s v_s (1+e) / (I_A + m_s L^2)
   D. w = e v_s / L
   Answer: A. Substituting v_s' = w L - e v_s into the angular-momentum equation gives m_s v_s L (1+e) = (I_A + m_s L^2) w, which is 3.97 rad/s here. (Key point: Combined result)

5. After finding w = 3.97 rad/s, how is the sphere's velocity obtained and what is it?
   A. v_s' = w L + e v_s = 8.77 m/s
   B. v_s' = w L - e v_s = 0.76 m/s in the original direction
   C. v_s' = e v_s = 4 m/s
   D. v_s' = -0.76 m/s
   Answer: B. From restitution, v_s' = 3.97(1.2) - 0.8(5) = 0.76 m/s, positive, so the sphere still moves forward but slower than the rod end. (Key point: Sphere velocity after impact)

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