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A uniform solid sphere of radius \(r\) is rolling on a smooth horizontal surface with velocity \(V\) and angular velocity \(\omega=(V=\omega r) .\) The sphere collides with a sharp edge on the wall as shown in figure. The coefficient of friction between the sphere and the edge \(\mu=1 / 5\). Just after the collision the angular velocity of the sphere becomes equal to zero. The linear velocity of the sphere just after the collision is equal to (1) \(V\) (2) \(\frac{V}{5}\) (3) \(\frac{3 V}{5}\) (4) \(\frac{V}{6}\)

Short Answer

Expert verified
The linear velocity of the sphere just after the collision is \( \frac{3V}{5} \). (Option 3)

Step by step solution

01

Understanding the Problem

The problem involves a sphere colliding with a wall. Initially, the sphere has a linear velocity \( V \) and angular velocity \( \omega = \frac{V}{r} \). The friction coefficient between the sphere and the wall is \( \mu = \frac{1}{5} \). After the collision, the sphere's angular velocity becomes zero, and we need to find its new linear velocity.
02

Apply Angular Impulse-Momentum Principle

For the sphere, the change in angular momentum can be calculated using the impulse-momentum principle. The initial angular momentum is \( I \omega \), where \( I = \frac{2}{5}mr^2 \) for a solid sphere. After the collision, the angular momentum is zero. The frictional impulse causes this change, calculated as \( -\mu mg \Delta t \cdot r = I \omega \).
03

Set Up Equations

Using the impulse angular momentum change, \( -\frac{1}{5}mg \Delta t \cdot r = \frac{2}{5}mr^2 \cdot \omega \). Simplifying the equation gives us \( -\frac{1}{5}mg \Delta t \cdot r = \frac{2}{5}mr \cdot V \), since \( \omega = \frac{V}{r} \).
04

Solve for Impulse Duration \(\Delta t\)

To find \( \Delta t \), solve the equation for \( \Delta t \), resulting in \( \Delta t = \frac{2r}{g} \). Plug this back in to find the linear velocity after collision using the impulse linear momentum change.
05

Calculate Change in Linear Velocity

The linear impulse-momentum relation is \( mg \Delta t \mu = m(V - v_{after}) \). Substitute \( \Delta t \) to find \( V - v_{after} = \frac{1}{5}g \cdot \frac{2r}{g} = \frac{2}{5} \cdot r \).
06

Solve for Final Linear Velocity

Since the initial velocity before the collision is \( V \) and \( V - v_{after} = \frac{2}{5}V \), the final velocity \( v_{after} = V - \frac{2}{5}V = \frac{3}{5}V \).
07

Conclusion

The sphere's linear velocity just after the collision is \( \frac{3}{5}V \). Therefore, the correct answer is option (3) \( \frac{3 V}{5} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Dynamics
Rotational dynamics involves the study of objects in rotation. For any rotating object, understanding how it moves and responds to forces is essential. In this scenario, the uniform solid sphere is rolling smoothly. This means both linear and angular velocities are in play. Coincidentally, for the sphere, these velocities relate through the equation \( V = \omega r \). This signifies a pure rolling motion. - **Moment of Inertia:** A central concept in rotational dynamics is the moment of inertia, \( I \). It's an object's resistance to changes in its rotational motion. For a solid sphere, like the one in this problem, \( I = \frac{2}{5}mr^2 \). When the sphere collides with the wall, all these factors influence how it stops rotating. The wall applies forces that interrupt both linear and rotational movements, showcasing the interplay between linear and angular momenta.
Friction
Friction is the resisting force from surfaces in contact. In this case, the friction between the sphere and the edge of the wall is crucial. It transforms rotational kinetic energy into heat, causing the sphere to stop spinning. - **Role of Friction:** Here, the given coefficient of friction \( \mu = \frac{1}{5} \) determines how much frictional force acts during the collision. - **Frictional Force:** It can be calculated as \( \mu mg \), with 'm' for mass and 'g' for gravity. This force affects how quickly the sphere loses its angular velocity.During the collision, the friction acts tangentially, applying torque that stops the sphere's rotation. Without this crucial frictional force, the sphere would continue spinning post-collision.
Impulse-Momentum Principle
The impulse-momentum principle connects changes in momentum to the impulses applied. In essence, any force applied over time (impulse) changes an object’s momentum, whether linear or angular. For angular momentum, the principle is expressed as: - **Equation:** \( I \omega_{initial} + \tau \Delta t = I \omega_{final} \), where \( \tau \) (torque) is from friction.- **Angular Impulse:** Specifically, this torque from friction decreases angular momentum to zero.In linear terms, the same impulse also alters the sphere's linear momentum. This computed change reveals the sphere's final velocity post-collision, illustrating that impulse impacts both movement types.
Collision
Collisions, like the one this sphere undergoes with the wall, significantly alter motion. Here, the initial rotational and linear velocities drop due to the collision's abrupt forces.- **Types of Collisions:** What we see is an inelastic collision. The sphere's angular velocity halts, suggesting internal kinetic energy loss, converted largely to heat due to friction.- **Change in Motion:** The friction force, working during \( \Delta t \), generates an impulse that changes linear velocity to \( \frac{3}{5}V \).This collision showcases energy shifts between rotational and linear forms. Ultimately, it highlights how initial conditions (like \( V = \omega r \)) can change dramatically upon impact, especially within multi-body interactions.

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Most popular questions from this chapter

A yo-yo is placed on a rough horizontal surface and a constant force \(F\), which is less than its weight, pulls it vertically. Due to this (1) friction force acts towards left, so it will move towards left (2) friction force acts towards right, so it will move towards right (3) it will move towards left, so friction force acts towards left (4) it will move towards right so friction force acts towards right

A ring of radius \(R\) is first rotated with an angular velocity \(\omega_{0}\) and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is \(\mu\). Time after which its angular speed is reduced to half is (I) \(\frac{\omega_{0} \mu R}{2 g}\) (2) \(\frac{\omega_{0} g}{2 \mu R}\) (3) \(\frac{2 \omega_{0} R}{\mu g}\) (4) \(\frac{\omega_{0} R}{2 \mu g}\)

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A uniform rod is resting freely over a smooth horizontal plane. A particle moving horizontally strikes at one end of the rod normally and gets stuck. Then (1) the momentum of the particle is shared between the particle and the rod and remains conserved (2) the angular momentum about the mid-point of the rod before and after the collision is equal (3) the angular momentum about the centre of mass of the combination before and after the collision is equal (4) the centre of mass of the rod particle system starts to move translationally with the original momentum of the particle

A circular platform is mounted on a vertical frictionless axle. Its radius is \(r=2 \mathrm{~m}\) and its moment of inertia is \(I=200 \mathrm{~kg} \mathrm{~m}^{2} .\) It is initially at rest. A \(70 \mathrm{~kg}\) man stands on the edge of the platform and begins to walk along the edge at speed \(v_{0}=10 \mathrm{~ms}^{-1}\) relative to the ground. The angular velocity of the platform is (1) \(1.2 \mathrm{rad} \mathrm{s}^{-1}\) (2) \(0.4 \mathrm{rad} \mathrm{s}^{-1}\) (3) \(2.0 \mathrm{rad} \mathrm{s}^{-1}\) (4) \(0.7 \mathrm{rad} \mathrm{s}^{-1}\)

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