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Rotational Motion question

2023 · 13 Apr · Shift 2 · Q64

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Rotational Motion
2023 · 13 Apr · Shift 2 · Q64

Rotational Motion question

2023 · 13 Apr · Shift 2 · Q64

JEE MainPhysicsRotational MotionNumerical+4 / −1

A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be

\underline{\hspace{2cm}}

rad s

^{-2}

.

Numerical answer

View written solutionFree

Correct answer: 15

Given data

Mass of hollow cylinder: $m = 5\,\text{kg}$
Radius: $R = 70\,\text{cm} = 0.70\,\text{m}$
Pulling force: $F = 52.5\,\text{N}$

We need the angular acceleration $\alpha$ .

Torque produced by the force

Since the rope is wound around the cylinder and pulled tangentially, the torque is

$\tau = F R$

So,

$\tau = 52.5 \times 0.70 = 36.75\,\text{N m}$

Moment of inertia of a hollow cylinder

For a hollow cylinder (thin cylindrical shell) about its axis,

$I = mR^2$

Thus,

$I = 5 \times (0.70)^2 = 5 \times 0.49 = 2.45\,\text{kg m}^2$

Use rotational equation of motion

$\tau = I\alpha$

Therefore,

$\alpha = \frac{\tau}{I} = \frac{36.75}{2.45} = 15\,\text{rad s}^{-2}$

Final answer

$\boxed{15}$

The angular acceleration of the cylinder is $15\,\text{rad s}^{-2}$ .

Comparison with stored correct answer

Stored correct answer = $15$

Our derived answer = $15$

So, the answer matches the stored correct answer.

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