---
title: "Ball strikes suspended rod, constrained motion"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:ball-strikes-suspended-rod-50n"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/ball-strikes-suspended-rod-50n
site: BCT Engineering
---

# Ball strikes suspended rod, constrained motion

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

- Asked in: 2081 Baishakh, 4+6 marks
- Video: [Moving Sphere strikes Rod resting on a Hinge](https://www.youtube.com/watch?v=Kq-ynicwQkE) — Indra Lal Subedi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/ball-strikes-suspended-rod-50n
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

Explain constrained plane motion with examples. The 50 N slender rod is suspended from the pin at A. If a 10 N ball B is thrown at the rod and strikes its centre with a horizontal velocity of 9 m/s, determine the angular velocity of the rod just after impact. The coefficient of restitution is $e = 0.4$. [Figure: vertical rod hanging from pin A, total length 1 m (0.5 m above and 0.5 m below its centre); ball B approaching horizontally at 9 m/s at the rod's centre.]

## Short answer

Constrained plane motion has kinematic relations linking linear and angular accelerations (e.g., fixed-axis rotation, rolling without slip). For the impact, angular momentum about pivot A is conserved, and using restitution e = 0.4 gives angular velocity of the rod just after impact 3.29 rad/s (ball rebounds at 1.96 m/s opposite direction).

## Answer

**Constrained plane motion** is plane motion of a rigid body in which the body is restricted by pins, guides, links or contact surfaces, so it cannot move freely. Because of these constraints, the acceleration of the mass centre $\bar a$ and the angular acceleration $\alpha$ are not independent: they are related by kinematic (geometric) equations. These relations, together with $\sum F = m\bar a$ and $\sum M_G = \bar I\alpha$, let us find the unknown accelerations and reactions.

**Examples**
- Rotation about a fixed pin not through $G$ (non-centroidal rotation), e.g. a rod or compound pendulum swinging about a pin: $a_t = r\alpha$, $a_n = r\omega^2$.
- A wheel or disc rolling without slipping: $\bar v = \omega r$ and $\bar a = \alpha r$.
- A connecting rod in a slider-crank mechanism, whose ends are guided by the crank pin and the piston.
- A ladder or rod sliding with its ends on a wall and the floor, or a collar sliding on a fixed rod.

#### Impact problem
The rod hangs from the pin $A$ and the ball strikes its centre $G$, $d = 0.5$ m below $A$. During the impact the weights are non-impulsive and the pin reaction passes through $A$, so neither gives a moment about $A$ and the angular momentum about $A$ is conserved.

#### Data
$$
m_r = \frac{50}{9.81} = 5.097\ \text{kg}, \qquad m_b = \frac{10}{9.81} = 1.019\ \text{kg}, \qquad I_A = \frac13 m_rL^2 = \frac13(5.097)(1)^2 = 1.699\ \text{kg m}^2
$$

Let $v_b'$ be the ball's velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

#### Conservation of angular momentum about A
$$
m_b v\,d = I_A\omega + m_b v_b'\,d \quad (1)
$$

$$
1.019(9)(0.5) = 1.699\,\omega + 1.019\,v_b'(0.5)
$$

#### Coefficient of restitution
The velocity of the rod at the point of impact is $\omega d$:

$$
e = \frac{\omega d - v_b'}{v - 0} \;\Rightarrow\; 0.5\omega - v_b' = 0.4(9) = 3.6 \quad (2)
$$

#### Solving
Substituting $v_b' = 0.5\omega - 3.6$ in (1):

$$
4.587 = 1.699\omega + 0.5097(0.5\omega - 3.6) \;\Rightarrow\; \omega = \frac{4.587 + 0.5097(3.6)}{1.699 + 0.5097(0.5)} = 3.29\ \text{rad/s}
$$

Then $v_b' = 0.5(3.29) - 3.6 = -1.96$ m/s.

**Answer: the angular velocity of the rod just after impact is $\omega = 3.29$ rad/s; the ball rebounds at $1.96$ m/s, opposite to its original direction ($v_b' = -1.96$ m/s).**

## Check yourself

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

1. What is constrained plane motion?
   A. Motion with no external forces
   B. Motion with constant velocity
   C. Motion where guides, pins, links or contact restrict the path, so coordinates are related by geometry
   D. Free motion in all directions
   Answer: C. In constrained motion the position is restricted, for example a piston in a cylinder, a collar on a rod or a wheel rolling without slipping. (Key point: Constrained motion)

2. Which one is an example of constrained motion with a kinematic relation between speeds?
   A. A projectile in flight
   B. A wheel rolling without slipping, with v = w r
   C. A freely falling stone
   D. A satellite far from any body
   Answer: B. Rolling without slipping ties the centre speed to the angular speed through v = w r. (Key point: Example of constraint)

3. Why is angular momentum about the pin A conserved during the ball-rod impact?
   A. The pin reaction passes through A and the weights are non-impulsive, so there is no moment about A
   B. Linear momentum is conserved, so angular momentum is also
   C. The impact is perfectly elastic
   D. The rod does not rotate
   Answer: A. The pin reaction acts through A and the weights are non-impulsive, so no impulsive moment acts about A and angular momentum of ball + rod is conserved. (Key point: Conservation principle in the impact)

4. Which equation expresses the coefficient of restitution for this impact (d = 0.5 m)?
   A. e = v_b' / v
   B. e = (w d - v_b') / v, so 0.5 w - v_b' = 3.6
   C. e = (w d + v_b') / v
   D. e = w / v
   Answer: B. e is the ratio of relative speed of separation (rod point speed w d minus ball speed v_b') to the approach speed v = 9 m/s. (Key point: Restitution equation)

5. What is the outcome after impact?
   A. w = 3.29 rad/s, and the ball rebounds at 1.96 m/s
   B. w = 3.29 rad/s, and the ball continues forward at 1.96 m/s
   C. w = 9 rad/s, and the ball stops
   D. w = 1.96 rad/s, and the ball rebounds at 3.29 m/s
   Answer: A. Solving the two equations gives w = 3.29 rad/s and v_b' = 0.5(3.29) - 3.6 = -1.96 m/s, meaning the ball reverses direction. (Key point: Result of impact)

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