Question:

A string is wound around a hollow cylinder of mass $5\, kg$ and radius $0.5\, m$. If the string is now pulled with a horizontal force of $40\, N$, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) :

Updated On: Sep 27, 2024
  • $12 \; rad/s^2$
  • $16 \; rad/s^2$
  • $10 \; rad/s^2$
  • $20 \; rad/s^2$
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The Correct Option is B

Solution and Explanation

$40 + f = m(R \alpha)$ .....(i)
$40 \times R - f \times R = mR^2 \alpha $
$40 - f = mR \alpha$ ...... (ii)
From (i) and (ii)
$\alpha = \frac{40}{mR} = 16 $
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System of Particles and Rotational Motion

  1. The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
  2. The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
  3. The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.