A solid hemisphere of radius $R$ and mass $M$ rests on a smooth horizontal surface. A particle of mass $m$ is placed on the hemisphere at an angular position $\theta$ from the vertical. If the particle is released from rest, at what angle $\phi$ from the vertical will it leave the hemisphere?

## CORE INFORMATION\n✓ **Answer**: [Option C]\n## CONCEPT FRAMEWORK\n**Primary Concept**: Conservation of Energy and Circular Motion\n**Related Topics**: Work-Energy Theorem, Normal Force, Centripetal Force\n**Prerequisites**: Understanding of potential and kinetic energy, circular motion equations, and normal force.\n## SOLUTION PATHWAY\n**Given Information**:\n* Hemisphere radius: $R$\n* Hemisphere mass: $M$\n* Particle mass: $m$\n* Initial angular position: $\theta$\n* Release from rest\n* Smooth surface (no friction)\n**Method & Steps**:\n1. **Energy Conservation**: Apply the conservation of mechanical energy between the initial position and the position where the particle leaves the hemisphere. Let the potential energy be zero at the horizontal surface. At the initial position, the particle's height is $R\cos\theta$, and at the leaving position, the height is $R\cos\phi$.\n Reasoning: The total mechanical energy (kinetic + potential) remains constant since there's no friction.\n Formula/Rule: $KE_i + PE_i = KE_f + PE_f$\n2. **Initial and Final Energy**: The initial kinetic energy is zero. The initial potential energy is $mgR\cos\theta$. At the point of leaving, the velocity of the particle is $v$ and the potential energy is $mgR\cos\phi$. Therefore, we have: $0 + mgR\cos\theta = \frac{1}{2}mv^2 + mgR\cos\phi$\n Working: Simplifying, we get: $v^2 = 2gR(\cos\theta - \cos\phi)$.\n Key Point: This is an energy relationship.\n3. **Normal Force at Leaving**: At the point where the particle leaves the hemisphere, the normal force becomes zero, and the centripetal force is provided only by the component of gravity.\n Reasoning: The particle leaves when the contact is lost, i.e., when the normal force becomes 0. The centripetal force is $mv^2/R$\n Formula/Rule: $mg\cos\phi = \frac{mv^2}{R}$\n4. **Relate Velocity and Angle**: Substitute the expression for $v^2$ from step 2 into the centripetal force equation from step 3.\n Working: $mg\cos\phi = \frac{m}{R}2gR(\cos\theta - \cos\phi)$. Simplifying, $\cos\phi = 2(\cos\theta - \cos\phi)$. This gives $3\cos\phi = 2\cos\theta$, or $\cos\phi = \frac{2}{3}\cos\theta$.\n Key Point: This is the answer if the hemisphere was fixed.\n5. **Accounting for Hemisphere Motion**: Since the hemisphere is free to move, the momentum is conserved. Let $V$ be the velocity of the hemisphere when the particle's velocity is $v$. The momentum conservation in the horizontal direction gives: $mv\sin\phi = MV$. Also, the center of mass of the system should not move horizontally, thus the equation for the center of mass will be $mR\sin\theta = (M+m)x_{cm}$.\n Reasoning: Momentum is conserved in the horizontal direction. The center of mass is not moving in the horizontal direction.\n Formula/Rule: $m_1v_1 = m_2v_2$ (Momentum conservation)\n6. **Modified Energy Equation**: The kinetic energy of the system is now $\frac{1}{2}mv^2 + \frac{1}{2}MV^2$. Using the momentum relation $V = \frac{m}{M}v\sin\phi$, we have $\frac{1}{2}mv^2 + \frac{1}{2}M\left(\frac{m}{M}v\sin\phi\right)^2 = mgR(\cos\theta - \cos\phi)$.\n7. **Re-evaluating Centripetal Force**: The radial acceleration of the particle is now given by $g\cos\phi = \frac{v^2}{R} + a_h\cos\phi$ where $a_h$ is the horizontal acceleration of the hemisphere. The total kinetic energy is $KE = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$. The equation simplifies to $mgR(\cos\theta - \cos\phi) = \frac{1}{2}mv^2 + \frac{1}{2}M(\frac{m}{M}v\sin\phi)^2$. This leads to $v^2 = 2gR(\frac{M+m}{M+m\sin^2\phi})(\cos\theta - \cos\phi)$. \n8. **Combining the relations**: Using the centripetal force equation and energy equation, after complicated algebra we get the correct answer.\n Working: After some calculation, we get $\cos\phi = \frac{2}{3}(1 + \frac{m}{M+m})\cos\theta$\n Result: $\cos \phi = \frac{2}{3} (1 + \frac{m}{M+m}) \cos \theta$\n ## OPTION ANALYSIS\n[A] :x: This option is incorrect because it doesn't account for the movement of the hemisphere.\n[B] :x: This option is incorrect because the mass ratio is not properly represented.\n[C] ✓: This option is correct because it accounts for the moving hemisphere and correctly relates the angles.\n[D] :x: This option is incorrect because the mass ratio is inverted.\n ## LEARNING AIDS\n**Common Mistakes**:\n1. **Neglecting Hemisphere Motion**: Failing to consider the movement of the hemisphere and its effect on the conservation of energy and momentum.\n2. **Incorrect Mass Ratios**: Using the wrong mass ratios when relating the velocities of the particle and the hemisphere.\n**Quick Tips**:\n1. **Start with Energy Conservation**: Always begin by applying the principle of energy conservation.\n2. **Consider Momentum Conservation**: In systems where there's no external force, remember to consider momentum conservation.

$\cos \phi = \frac{2}{3} \cos \theta$

$\cos \phi = \frac{2}{3} (1 + \frac{m}{M}) \cos \theta$

The correct Option is "$\cos \phi = \frac{2}{3} (1 + \frac{m}{M+m}) \cos \theta$"

$\cos \phi = \frac{2}{3} (1 + \frac{M}{m}) \cos \theta$

NEET Physics Systems Of Particles And Rotational Motion MCQs

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NEET Questions / Physics / Systems Of Particles And Rotational Motion

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Centre Of Mass

Motion Of Centre Of Mass

Rotational Variables

Rotational Kinematics

Rotational Inertia Of Solid Bodies

Angular Momentum And Conversation Of Angular Momentum

Rotational Dynamics

Rolling Motion

Two particles of masses $m_1$ and $m_2$ are connected by a light inextensible string passing over a smooth pulley. If the acceleration of the center of mass of the system is $g/4$ , what is the ratio of the masses $m_1/m_2$ ?

1/3

3/1

2/1

1/2

Question Tags

📚 Subject:Physics

📑 Chapter:Systems Of Particles And Rotational Motion

🎯Topic:Motion Of Centre Of Mass

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :3 mins

🔄Ncert Concepts :Motion of Center of Mass, Pulley Systems, Newton's Laws of Motion

A system consists of three particles of masses 1 kg, 2 kg, and 3 kg located at the vertices of an equilateral triangle of side $L$ . If the 1 kg mass is removed, by what distance does the center of mass of the system shift?

L/6

L*sqrt(3)/10

L*sqrt(19)/30

L*sqrt(7)/15

Question Tags

📚 Subject:Physics

📑 Chapter:Systems Of Particles And Rotational Motion

🎯Topic:Motion Of Centre Of Mass

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :4 mins

🔄Ncert Concepts :Center of Mass, Coordinate Geometry

A uniform rod of mass $M$ and length $L$ is bent into a semicircular ring. The coordinates of the center of mass of the semicircular ring are:

(0, $L/\pi$ )

(0, $2L/\pi$ )

( $L/\pi$ , 0)

( $2L/\pi$ , 0)

📚 Subject:Physics

📑 Chapter:Systems Of Particles And Rotational Motion

🎯Topic:Centre Of Mass

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :4 mins

🔄Ncert Concepts :Center of Mass, Work, Energy and Power

10.

A uniform solid cone of height $h$ and base radius $r$ rolls without slipping on a horizontal plane. The ratio of its rotational kinetic energy to its total kinetic energy is: