---
title: "Bullet embedded in bar, h = 260 mm"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:bullet-slender-bar-h260"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-slender-bar-h260
site: BCT Engineering
---

# Bullet embedded in bar, h = 260 mm

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

- Asked in: 2078 Kartik, 6 marks
- Video: [Bullet embedded on a panel | Applied Mechanics Dynamic | Plane motion of Rigid Bodies | Unit:9 |PU|](https://www.youtube.com/watch?v=U5xO8ZO4sKk) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-slender-bar-h260
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

A bullet weighing 40 gm is fired with horizontal velocity of 600 m/s into the lower end of a slender 7 kg bar of length L = 600 mm. Knowing that h = 260 mm and that the bar is initially at rest, determine (a) the angular velocity of bar immediately after the bullet becomes embedded, (b) the impulsive reaction at C, assuming that the bullet becomes embedded in 0.001 s. [Figure: vertical bar AB of length L pivoted at C, distance h below A; bullet with velocity $v_0$ strikes the lower end B.]

## Short answer

As with h = 240 mm, conserve angular momentum about C, then apply impulse–momentum. With h = 260 mm, the distances d_G and d_B change; results: ω = 36.13 rad/s and the impulsive reaction at C is approximately 13391 N opposite to the bullet velocity.

## Answer

Angular momentum of the bar and bullet about the pivot $C$ is conserved during the impact (the reaction at $C$ has no moment about $C$; the weights are non-impulsive). Take the bullet mass $m_b = 0.040$ kg, $v_0 = 600$ m/s, bar mass $M = 7$ kg, $L = 0.600$ m, $h = 0.260$ m (distance of the pivot $C$ below $A$); the bullet enters at the lower end $B$.

```text
     A  o
        |   G = centre of bar
   h    |
     C  o-- pivot
        |
     B  o<--- bullet v0
```

#### Distances from C
- Centre of bar $G$ from $C$: $d_G = L/2 - h = 0.300 - 0.260 = 0.040$ m (below $C$)
- Bullet point $B$ from $C$: $d_B = L - h = 0.600 - 0.260 = 0.340$ m

#### Moment of inertia of the bar about C
$$
\bar I = \frac{1}{12}ML^2 = \frac{1}{12}(7)(0.6)^2 = 0.210\ \text{kg m}^2
$$

$$
I_C = \bar I + Md_G^2 = 0.210 + 7(0.040)^2 = 0.2212\ \text{kg m}^2
$$

#### (a) Angular velocity just after impact
The bullet is embedded, so its speed after impact is $\omega d_B$. Conservation of angular momentum about $C$:

$$
m_bv_0d_B = I_C\omega + m_b(\omega d_B)d_B
$$

$$
\omega = \frac{m_bv_0d_B}{I_C + m_bd_B^2} = \frac{0.04\times600\times0.340}{0.2212 + 0.04(0.340)^2} = 36.13\ \text{rad/s}
$$

#### (b) Impulsive reaction at C
Take the bullet direction as positive. Linear impulse-momentum for the system (bar + bullet), the impulse of the reaction being $C\,\Delta t$:

$$
m_bv_0 + C\,\Delta t = M(\omega d_G) + m_b(\omega d_B)
$$

$$
C\,\Delta t = 7(36.13)(0.040) + 0.04(36.13)(0.340) - 0.04(600) = -13.39\ \text{N s}
$$

$$
C = \frac{C\,\Delta t}{\Delta t} = \frac{-13.39}{0.001} = -13391\ \text{N}
$$

The negative sign means the reaction at $C$ acts opposite to the bullet's direction.

**Answer: (a) $\omega = 36.13$ rad/s; (b) $C \approx 13391$ N ($\approx 13.39$ kN) opposite to the bullet velocity.**

## Check yourself

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

1. Why is angular momentum about C used for the bullet-bar system?
   A. C is the mass centre
   B. The reaction at C has no moment about C and weights are non-impulsive
   C. The bullet passes through C
   D. Energy is conserved
   Answer: B. No external impulsive moment acts about C, so angular momentum about C is conserved. (Key point: Conservation about C)

2. With h = 0.26 m, I_C of the bar is:
   A. 0.210 kg m^2
   B. 0.2212 kg m^2
   C. 0.2352 kg m^2
   D. 0.0112 kg m^2
   Answer: B. I_C = 0.210 + 7(0.04)^2 = 0.2212 kg m^2, using d_G = 0.300 - 0.260 = 0.040. (Key point: Moment of inertia)

3. omega = m_b v0 d_B / (I_C + m_b d_B^2) with d_B = 0.34 m gives:
   A. 36.13 rad/s
   B. 35.94 rad/s
   C. 72 rad/s
   D. 24 rad/s
   Answer: A. 0.04x600x0.34 = 8.16; denominator 0.2212 + 0.004624 = 0.2258; omega = 36.13 rad/s. (Key point: Angular velocity)

4. C dt = -13.39 N s means:
   A. C acts in the direction of the bullet
   B. C acts opposite to the bullet, about 13.4 kN
   C. C is zero
   D. C is 13.39 N
   Answer: B. Negative sign: opposite to the bullet; C = 13.39 N s / 0.001 s = 13391 N. (Key point: Sign of reaction)

## Close variants set in other exams

- [Bullet embedded in bar, h = 240 mm](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Abullet-slender-bar-h240) — Applied Mechanics (Dynamics), chapter 6, asked 2×

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