---
title: "Collar on arm: velocity and relative acceleration"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:collar-on-rotating-arm-velocity-relative"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/collar-on-rotating-arm-velocity-relative
site: BCT Engineering
---

# Collar on arm: velocity and relative acceleration

Applied Mechanics (Dynamics) · Chapter 1: Curvilinear Motion of Particles

- Asked in: 2073 Shrawan, 6 marks
- Video: [Rigid Bodies Relative Motion Analysis: Acceleration Dynamics (step by step)](https://www.youtube.com/watch?v=ee_LrUTVE5o) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/collar-on-rotating-arm-velocity-relative
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

Rotation of the arm about O is defined by $\theta = 0.75t^2$ where $\theta$ is in radians and $t$ in seconds. Collar B slides along the arm such that $r = 1 - 0.3t^2$ where $r$ is in meters. After the arm has rotated through $45^\circ$, determine (a) the total velocity of the collar, (b) the total acceleration of the collar and (c) the relative acceleration of the collar with respect to the arm. [Figure: arm OA rotating about O at angle $\theta$, collar B at distance $r$ from O.]

## Short answer

At θ=45°, t=1.023 s. v_r=−0.614 m/s, v_θ=1.053 m/s, total v=1.219 m/s. a_r=−2.216 m/s², a_θ=−0.856, total a=2.376 m/s². Relative acceleration a_rel = r̈ = −0.6 m/s², toward O.

## Answer

#### Time when $\theta=45^\circ$
$$
\theta=45^\circ=0.7854\ \text{rad}=0.75t^{2}\ \Rightarrow\ t^{2}=1.0472,\ t=1.0233\ \text{s}
$$

#### Quantities at this instant
$$
\begin{aligned}
r&=1-0.3t^{2}=0.6858\ \text{m}\\
\dot r&=-0.6t=-0.6140\ \text{m/s},\qquad \ddot r=-0.6\ \text{m/s}^2\\
\dot\theta&=1.5t=1.5350\ \text{rad/s},\qquad \ddot\theta=1.5\ \text{rad/s}^2
\end{aligned}
$$

#### (a) Total velocity
$$
\begin{aligned}
v_r&=\dot r=-0.6140\ \text{m/s}\\
v_\theta&=r\dot\theta=0.6858(1.5350)=1.0528\ \text{m/s}\\
v&=\sqrt{v_r^{2}+v_\theta^{2}}=1.219\ \text{m/s}
\end{aligned}
$$

The velocity makes $120.3^\circ$ with $\hat e_r$, that is, $59.8^\circ$ with the direction along the arm towards O.

#### (b) Total acceleration
$$
\begin{aligned}
a_r&=\ddot r-r\dot\theta^{2}=-0.6-0.6858(1.5350)^{2}=-2.216\ \text{m/s}^2\\
a_\theta&=r\ddot\theta+2\dot r\dot\theta=0.6858(1.5)+2(-0.6140)(1.5350)=-0.856\ \text{m/s}^2\\
a&=\sqrt{2.216^{2}+0.856^{2}}=2.376\ \text{m/s}^2
\end{aligned}
$$

#### (c) Acceleration relative to the arm
The collar moves along the arm, so relative to the arm (a rotating frame) its motion is rectilinear along $r$:

$$
a_{rel}=\ddot r=-0.6\ \text{m/s}^2
$$

That is, $0.6$ m/s$^2$ directed towards O along the arm.

**Answer: (a) $v=1.219$ m/s; (b) $a=2.376$ m/s$^2$; (c) $a_{rel}=0.6$ m/s$^2$ towards O.**

## Check yourself

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

1. Why is theta converted to radians (0.7854 rad) before solving?
   A. Because theta = 0.75t^2 is a formula in radians
   B. Because time is in radians
   C. Because r is in radians
   D. No reason; degrees would give the same t
   Answer: A. The given law has theta in radians, so t^2 = 0.7854/0.75 = 1.047. (Key point: Time at 45 degrees)

2. How is the total velocity found?
   A. v = r_dot + r theta_dot
   B. v = sqrt(r_dot^2 + (r theta_dot)^2)
   C. v = r theta_double_dot
   D. v = sqrt(r_double_dot^2 + (r theta_dot)^2)
   Answer: B. Components are v_r = r_dot = -0.614 and v_theta = r theta_dot = 1.053 m/s, giving 1.219 m/s. (Key point: Total velocity)

3. Which formula gives a_theta?
   A. r theta_double_dot + 2 r_dot theta_dot
   B. r theta_double_dot - 2 r_dot theta_dot
   C. r theta_dot^2 + 2 r_dot
   D. r_double_dot + 2 r theta_dot
   Answer: A. a_theta = 0.6858(1.5) + 2(-0.614)(1.535) = -0.856 m/s^2. (Key point: Transverse acceleration)

4. What is the acceleration of the collar relative to the arm?
   A. r theta_double_dot = 1.03 m/s^2
   B. r_double_dot = -0.6 m/s^2, i.e. 0.6 m/s^2 towards O along the arm
   C. The total acceleration 2.376 m/s^2
   D. Zero, because the collar moves with the arm
   Answer: B. Relative to the arm the collar moves in a straight line along r, so a_rel = r_double_dot. (Key point: Relative acceleration)

## Close variants set in other exams

- [Collar sliding on rotating arm, acceleration](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-on-rotating-arm-acceleration) — Applied Mechanics (Dynamics), chapter 1, asked 2×