/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 A particle of mass \(m\) is plac... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 94

A particle of mass \(m\) is placed on top of a smooth hemispherical wedge of mass \(4 m\) at rest. The hemispherical wedge is kept on a frictionless horizontal surface. The particle is given a gentle push. The angular velocity of the particle relative to the wedge if wedge has velocity \(v\) when particle has fallen an angle \(\theta\) with respect to the wedge is (A) \(\frac{5 v}{R \cos \theta}\) (B) \(\frac{v}{R \cos \theta}\) (C) \(\frac{4 v}{R \cos \theta}\) (D) \(\frac{5 v}{R \sin \theta}\)

Short Answer

Expert verified
The angular velocity of the particle relative to the wedge, when the wedge has velocity \(v\) and the particle has fallen at an angle \( \theta \) is given by, \( \frac{4v}{R \cos\theta} \). Therefore, the correct answer is (C)

Step by step solution

01

Conservation of Linear Momentum

Since there's no external force acting on the system (the particle and the hemisphere), the linear momentum of the system is conserved. Therefore: \( Mv = mv_{cm}, \) where \( M = 4m, \) \( m = \) mass of the particle, \( v = \) velocity of the hemisphere, and \( v_{cm} = \) velocity of the particle's center of mass with respect to the ground.
02

Deriving Velocity of particle's center of mass

From the above equation, let's derive expression for velocity of the particle鈥檚 center of mass (which is at a perpendicular distance of Rcos胃 from the center of the hemisphere). \( v_{cm} = \frac{Mv}{m} = \frac{4 v}{1}\)
03

Derive Angular velocity

Angular velocity of the particle (w) relative to the hemisphere can be obtained by dividing the velocity of the particle's center of mass by the radius (which is \(R \cos\theta\)). So we have: \( w = \frac{v_{cm}}{R \cos \theta} = \frac{4v}{R \cos \theta} \). So, the correct answer is (C) \( \frac{4v}{R \cos \theta} \)

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