Question

23. 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 30N?​

Answer 1

Answer:

10m/s² is the angular acceleration .

Explanation:

The moment of inertia of the hollow cylinder about its geometric axis:

I=mr

2

=3×0.4

2

kgm

2

The Torque developed is given by:

τ=rF=0.4×30Nm=12Nm

We know that the relation between the torque and angular acceleration is:

τ=Iα

⟹α=25rads

−2

Thus linear acceleration(a)=rα=0.4×25m/s²

=10m/s²

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Answer 2

Given :-

Mass of the hollow cylinder = 3 kg

Radius of the hollow cylinder = 40 cm

Force applied by pulling the rope = 30 N

To Find :-

The  angular acceleration of the cylinder.

Solution :-

We know that,

• m = Mass
• i = Momentum of inertia
• t = Torque
• f = Force
• r = Radius

Using the formula,

$\underline{\boxed{\sf Momentum \ of \ inertia=mr^2}}$

Given that,

Mass (m) = 3 kg

Radius (r) = 40 cm = 0.4 m

Substituting their values,

⇒ 3 × (0.4)²

⇒ 3 × 0.16

⇒ 0.48 kg m²

Using the formula,

$\underline{\boxed{\sf Torque=Force \times Radius}}$

Given that,

Force (f) = 30 N

Radius (r) = 0.4 m

Substituting their values,

⇒ t = 30 × 0.4

⇒ t = 12 Nm

Using the formula,

$\underline{\boxed{\sf Torque=Momentum \ of \ inertia \times Angular \ acceleration}}$

Given that,

Torque (t) = 12 Nm

Momentum of inertia (i) = 0.48 kg m²

Substituting their values,

⇒ τ = Iα

⇒ 12 = 0.48 × α

⇒ α = 12/0.48

⇒ α = 25 rad s⁻²

Therefore, the angular acceleration of the cylinder is 25 rad s⁻².