---
title: "Wheel pulled by cord on inner drum, 45 kg"
subject: "Applied Mechanics (Dynamics)"
chapter: 5
key: "semester_3/applied_mechanics_dynamics:wheel-inner-drum-45kg"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/wheel-inner-drum-45kg
site: BCT Engineering
---

# Wheel pulled by cord on inner drum, 45 kg

Applied Mechanics (Dynamics) · Chapter 5: Plane Motion of Rigid Bodies: Forces, Moments, and Accelerations

- Asked in: 2078 Kartik, 8 marks
- Video: [A cord is wrapped around the inner drum of a wheel and pull horizontally with force 200N](https://www.youtube.com/watch?v=uVJ7zkIVLPU) — Civil Engg Network
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/wheel-inner-drum-45kg
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5

## Question

A cord is wrapped around the inner drum of a wheel and pulled horizontally with a force of 200 N. The wheel has a mass of 45 kg and radius of gyration of 70 mm. Knowing that $\mu_s = 0.2$ and $\mu_k = 0.15$, determine the acceleration of G and angular acceleration of wheel. [Figure: wheel with outer radius $r = 100$ mm and inner drum radius $r = 60$ mm; 200 N pulled horizontally to the right from the inner drum; wheel on rough ground.]

## Short answer

Assume rolling without slipping to find required friction. Since it exceeds μ_s N, the wheel slips; use kinetic friction F_k = μ_k N. Then a_G = (P − F_k)/m = 2.97 m/s² to the right; α = (P r − F_k R)/¯ I = 24.39 rad/s² counter‑clockwise.

## Answer

**Assumptions.** The cord leaves the bottom of the inner drum (radius $r=0.060$ m) and pulls horizontally to the right; the wheel (outer radius $R=0.100$ m) rests on rough ground. Take $x$ to the right; the sense of $\alpha$ is stated in each step. The friction force $F$ acts at the contact point.

#### Data
$m = 45$ kg, $\bar k = 0.070$ m, $P = 200$ N, $\mu_s = 0.20$, $\mu_k = 0.15$

$$
\bar I = m\bar k^2 = 45(0.070)^2 = 0.2205\ \text{kg m}^2, \qquad N = mg = 45\times9.81 = 441.45\ \text{N}
$$

#### Step 1: assume rolling without slipping
Rolling to the right means the wheel turns clockwise, so $\bar a = \alpha R$ with $\alpha$ clockwise. The pull $P$ (below $G$) gives a counter-clockwise moment $Pr$ about $G$; friction $F$, taken to the left at the contact point, gives a clockwise moment $FR$:

$$
\begin{aligned}
P - F &= m\bar a\\
FR - Pr &= \bar I\alpha
\end{aligned}
$$

Eliminating $F$ with $\bar a=\alpha R$:

$$
\alpha = \frac{P(R-r)}{m(\bar k^2 + R^2)} = \frac{200(0.100-0.060)}{45(0.0049+0.0100)} = 11.93\ \text{rad/s}^2\ \text{(clockwise)}, \qquad \bar a = \alpha R = 1.19\ \text{m/s}^2
$$

Friction required: $F = P - m\bar a = 200 - 45(1.19) = 146.31$ N.

Maximum available: $F_{max} = \mu_s N = 0.20\times441.45 = 88.29$ N.

Since $F = 146.31\ \text{N} > F_{max}$, the wheel **slips**. The assumption is wrong, and kinetic friction acts: $F = \mu_k N = 0.15\times441.45 = 66.22$ N.

#### Step 2: sliding (kinetic friction)
$$
\begin{aligned}
\bar a &= \frac{P - F_k}{m} = \frac{200 - 66.22}{45} = 2.97\ \text{m/s}^2\ (\to)\\
\alpha &= \frac{Pr - F_kR}{\bar I} = \frac{200(0.060) - 66.22(0.100)}{0.2205} = 24.39\ \text{rad/s}^2\ \text{(counter-clockwise)}
\end{aligned}
$$

With $\alpha$ counter-clockwise, the contact point accelerates forward relative to the ground ($\bar a + \alpha R>0$), so it slips forward. This confirms that friction acts to the left.

**Answer: $\bar a_G = 2.97$ m/s² to the right; $\alpha = 24.39$ rad/s² counter-clockwise.**

## Check yourself

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

1. Assuming rolling without slipping, which relation links a and alpha?
   A. a = alpha r (inner radius)
   B. a = alpha R (outer radius)
   C. a = alpha k
   D. a = 2 alpha R
   Answer: B. The contact point at radius R is the instantaneous centre, so a = alpha R. (Key point: Rolling assumption)

2. Required friction 146.31 N vs. mu_s N = 88.29 N. The conclusion is:
   A. No slipping, a = 1.19 m/s^2
   B. The wheel slips and friction is mu_k N = 66.22 N
   C. The wheel slips and friction is 88.29 N
   D. Friction equals 146.31 N
   Answer: B. Required friction exceeds F_max, so the assumption fails and kinetic friction acts. (Key point: Slip check)

3. With k = 0.070 m and m = 45 kg, the moment of inertia is:
   A. 0.2205 kg m^2
   B. 3.15 kg m^2
   C. 0.0049 kg m^2
   D. 0.315 kg m^2
   Answer: A. I = m k^2 = 45 x 0.0049 = 0.2205 kg m^2. (Key point: Radius of gyration)

4. With F_k = 66.22 N, the acceleration of G and alpha are:
   A. 2.97 m/s^2 and 24.39 rad/s^2
   B. 1.19 m/s^2 and 11.93 rad/s^2
   C. 2.97 m/s^2 and 11.93 rad/s^2
   D. 4.44 m/s^2 and 24.39 rad/s^2
   Answer: A. a = (200 - 66.22)/45 = 2.97; alpha = (12 - 6.622)/0.2205 = 24.39 rad/s^2. (Key point: Final results)

## Close variants set in other exams

- [Wheel pulled by cord on inner drum, 50 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Awheel-inner-drum-50kg) — Applied Mechanics (Dynamics), chapter 5, asked 2×

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