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

# Collar sliding on rotating arm, acceleration

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

- Asked in: 2079 Baishakh, 4 marks
- Video: [Dynamics 12.169 - The slotted link is pinned at O, and as a result of the constant angular velocity](https://www.youtube.com/watch?v=oN1e1C5hXHk) — Learning by Teaching
- 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-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

Rotation of the arm about O is defined by $\theta = 0.23t^2$ where $\theta$ is in radians and $t$ in seconds. Collar B slides along the arm such that $r = 0.9 - 0.12t^2$ where $r$ is in meters. After the arm has rotated through $35^\circ$, determine the total acceleration of the collar. [Figure: arm OA rotating about O at angle $\theta$, collar B at distance $r$ from O.]

## Short answer

θ=35° gives t=1.63 s. Then r=0.581 m, ṙ=−0.391 m/s, r̈=−0.24 m/s², θ̇=0.7497 rad/s, θ̈=0.46. a_r = r̈−rθ̇² = −0.567 m/s²; a_θ = rθ̈+2ṙθ̇ = −0.319 m/s²; total a = 0.650 m/s².

## Answer

Use polar coordinates with $r$ along the arm and $\theta$ the arm rotation.

#### Time when $\theta=35^\circ$
$$
\theta=35^\circ=0.6109\ \text{rad}=0.23t^{2}\ \Rightarrow\ t^{2}=2.656,\ t=1.6297\ \text{s}
$$

#### Kinematic quantities at this instant
$$
\begin{aligned}
r&=0.9-0.12t^{2}=0.9-0.12(2.656)=0.5813\ \text{m}\\
\dot r&=-0.24t=-0.3911\ \text{m/s},\qquad \ddot r=-0.24\ \text{m/s}^2\\
\dot\theta&=0.46t=0.7497\ \text{rad/s},\qquad \ddot\theta=0.46\ \text{rad/s}^2
\end{aligned}
$$

#### Acceleration components
$$
\begin{aligned}
a_r&=\ddot r-r\dot\theta^{2}=-0.24-0.5813(0.7497)^{2}=-0.5667\ \text{m/s}^2\\
a_\theta&=r\ddot\theta+2\dot r\dot\theta=0.5813(0.46)+2(-0.3911)(0.7497)=-0.3190\ \text{m/s}^2\\
a&=\sqrt{a_r^{2}+a_\theta^{2}}=0.650\ \text{m/s}^2
\end{aligned}
$$

Both components are negative: the acceleration points towards O and against the sense of increasing $\theta$, at $\tan^{-1}(0.319/0.567)=29.4^\circ$ from the line towards O.

**Answer: $a_r=-0.567$ m/s$^2$, $a_\theta=-0.319$ m/s$^2$, total $a\approx0.650$ m/s$^2$.**

## Check yourself

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

1. How is the time found when the arm has rotated 35 degrees?
   A. Set 35 = 0.23t^2 directly
   B. Convert 35 degrees to 0.6109 rad and solve 0.23t^2 = 0.6109
   C. Set r = 35
   D. Differentiate theta and set it to 35
   Answer: B. theta must be in radians; t^2 = 0.6109/0.23 = 2.656, t = 1.63 s. (Key point: Finding the time)

2. Which set of quantities is needed to compute a_r and a_theta?
   A. r, r_dot, r_double_dot, theta_dot, theta_double_dot
   B. Only r and theta
   C. Only theta_dot and r
   D. r_dot and theta only
   Answer: A. a_r and a_theta depend on r, its first two derivatives and the first two derivatives of theta. (Key point: Rates needed)

3. Which expression gives a_r here?
   A. r_double_dot - r theta_dot^2
   B. r_double_dot + r theta_dot^2
   C. r theta_double_dot + 2 r_dot theta_dot
   D. r_dot - r theta_dot
   Answer: A. a_r = -0.24 - 0.5813(0.7497)^2 = -0.567 m/s^2. (Key point: Radial component)

4. Both a_r and a_theta are negative. What does this mean?
   A. The collar is decelerating to rest
   B. Acceleration points towards O and against increasing theta
   C. Acceleration points away from O
   D. The calculation must be wrong since magnitudes cannot be negative
   Answer: B. Negative a_r means towards O; negative a_theta means opposite to the direction of increasing theta. Magnitude is sqrt(0.567^2+0.319^2) = 0.650 m/s^2. (Key point: Meaning of negative components)

## Close variants set in other exams

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