Skip to content
Exercises · 6.13

Q.A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

Jharkhand JacTextbookSubjective· 3mImportance★★★★★est
44% · 25/57 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The rope's pull creates a torque that rotates the cylinder; using τ=Iα\tau = I\alpha and the no-slip condition a=αRa = \alpha R, we find α=25 rad/s2\alpha = 25\ \text{rad/s}^2 and a=10 m/s2a = 10\ \text{m/s}^2.

The key here is to connect the force on the rope to the rotational motion of the cylinder. A hollow cylinder (like a thin-walled pipe) has all its mass concentrated at the rim, so its moment of inertia is simply MR2MR^2. That’s the first thing to lock in — if this were a solid cylinder, the inertia would be 12MR2\frac12 MR^2, and the answer would change.

The rope is wound around the cylinder and pulled tangentially. Since there’s no slipping, the rope’s linear acceleration equals the tangential acceleration of a point on the cylinder’s surface. That’s the bridge between translation and rotation.

Let’s work it through.

  1. Moment of inertia of the hollow cylinder For a hollow cylinder (or a thin ring) rotating about its central axis:

I=MR2I = MR^2

Here M=3 kgM = 3\ \text{kg} and R=40 cm=0.4 mR = 40\ \text{cm} = 0.4\ \text{m}.

I=3×(0.4)2=3×0.16=0.48 kg m2I = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48\ \text{kg m}^2

  1. Torque due to the rope The force of 30 N acts tangentially at the rim, so the lever arm is exactly RR.

τ=F⋅R=30×0.4=12 N m\tau = F \cdot R = 30 \times 0.4 = 12\ \text{N m}

  1. Angular acceleration from Newton’s second law for rotation

τ=Iα⇒α=τI=120.48\tau = I \alpha \quad \Rightarrow \quad \alpha = \frac{\tau}{I} = \frac{12}{0.48}

α=25 rad/s2\alpha = 25\ \text{rad/s}^2

Tip

Notice that 0.48=48100=12250.48 = \frac{48}{100} = \frac{12}{25}, so 120.48=12×2512=25\frac{12}{0.48} = 12 \times \frac{25}{12} = 25. This kind of cancellation often appears in such problems — keep an eye on fractions.

  1. Linear acceleration of the rope …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.