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Step-by-Step Solution
Step 1: Identify the Physical Situation
A rope is wound around a hollow cylinder (ring). When the rope is pulled with a force, a torque is exerted on the cylinder, causing it to rotate. We need to find its angular acceleration.
Step 2: Write Down Known Quantities
Mass of the hollow cylinder, $M = 3 \, \text{kg}$
Radius of the hollow cylinder, $R = 0.40 \, \text{m}$
Applied force, $F = 30 \, \text{N}$
Step 3: Express the Moment of Inertia for a Hollow Cylinder
For a hollow cylinder (or ring), the moment of inertia about its central axis is given by:
$$ I = M R^2. $$
Substituting the values:
$$ I = 3 \times (0.40)^2 = 3 \times 0.16 = 0.48 \, \text{kg·m}^2. $$
Step 4: Calculate the Torque
Torque ($\tau$) is given by the product of the force and the perpendicular distance from the axis of rotation:
$$ \tau = F \times R. $$
So,
$$ \tau = 30 \, \text{N} \times 0.40 \, \text{m} = 12 \, \text{N·m}. $$
Step 5: Determine the Angular Acceleration
The relationship between torque, moment of inertia, and angular acceleration ($\alpha$) is:
$$ \tau = I \alpha. $$
Thus,
$$ \alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25 \, \text{rad·s}^{-2}. $$
Step 6: Final Answer
The angular acceleration of the cylinder is
$$ 25 \, \text{rad·s}^{-2}. $$
