Question:

A disc of mass 1 kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc of fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, it angular speed will be \(4\sqrt {\frac {x}{3R}}\) rad s−1 where x = ______ .(g = 10 ms–2)

Updated On: Mar 19, 2025
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Correct Answer: 5

Solution and Explanation

A disc of mass 1 kg and radius R is free to rotate about a horizontal axis
Loss in P.E = Gain in K.E.
\(2mgR=\frac 12[\frac 12mR^2+mR^2]w^2\)
\(2mgR=\frac 12×\frac 32mR^2w^2\)
\(w^2=\frac {8g}{3R}\)
\(w=\sqrt {\frac {8g}{3R}}\)
\(w=4\sqrt {\frac {g}{2×3R}}\)
\(⇒x=\frac g2\)
\(⇒x=5\)
So, the answer is \(5\).
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Concepts Used:

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.