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Q6.13:
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.

Solution :

Step 1: Given Variables & Initial Setup

To determine the angular and linear acceleration of the hollow cylinder, we identify the physical parameters provided:

  • Mass of the cylinder ($M$): $3\text{ kg}$
  • Radius of the cylinder ($R$): $40\text{ cm} = 0.4\text{ m}$
  • Applied Force ($F$): $30\text{ N}$
  • Moment of Inertia of a hollow cylinder about its central axis ($I$): $MR^2$ [Per the definition of the moment of inertia for a thin cylindrical shell]

Step 2: Calculating Moment of Inertia

Using the formula for the moment of inertia of a hollow cylinder:

$I = MR^2$

$I = 3\text{ kg} \times (0.4\text{ m})^2$

$I = 3 \times 0.16 = 0.48\text{ kg}\cdot\text{m}^2$

Step 3: Calculating Torque

Torque ($\tau$) is defined as the product of the force and the perpendicular distance from the axis of rotation:

$\tau = F \times R$ [Per the definition of torque]

$\tau = 30\text{ N} \times 0.4\text{ m} = 12\text{ N}\cdot\text{m}$

Step 4: Determining Angular Acceleration ($\alpha$)

Using Newton's Second Law for rotation, where $\tau = I\alpha$:

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{12\text{ N}\cdot\text{m}}{0.48\text{ kg}\cdot\text{m}^2}$

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

Step 5: Determining Linear Acceleration ($a$)

Since there is no slipping between the rope and the cylinder, the linear acceleration of the rope is equal to the tangential acceleration of the rim of the cylinder:

$a = R\alpha$ [Kinematic relationship for rotational motion]

$a = 0.4\text{ m} \times 25\text{ rad/s}^2$

$a = 10\text{ m/s}^2$

R = 0.4m F = 30N Axis

Final Solution: The angular acceleration of the cylinder is $25\text{ rad/s}^2$ and the linear acceleration of the rope is $10\text{ m/s}^2$.


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