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Sphere \(A\) of mass \(m\) and radius \(r\) rolls without slipping with a velocity \(\bar{v}_{1}\) on a horizontal surface when it hits squarely an identical sphere \(B\) that is at rest. Denoting by \(\mu_{k}\) the coefficient of kinetic friction between the spheres and the surface, neglecting friction between the spheres, and assuming perfectly elastic impact, determine \((a)\) the linear and angular velocities of each sphere immediately after the impact, (b) the velocity of each sphere after it has started rolling uniformly.

Short Answer

Expert verified
Sphere A and B have linear velocities of \( \bar{v}_1 \) and 0, respectively, immediately after impact. After rolling without slipping, \( \bar{v}_A = \bar{v}_B = \frac{2v_1}{2 + \mu_k \frac{g}{r} t} \). Sphere A's angular velocity \( \omega = \frac{\bar{v}_{A}}{r} \).

Step by step solution

01

Understand the Impact

Since the impact is perfectly elastic and assuming no friction between the spheres, we can use the coefficient of restitution, which is 1, for the impact between the spheres. This implies that the relative velocity after impact will be the negative of that before impact when considering the line of impact.
02

Use Conservation of Linear Momentum

The total linear momentum before and after the collision must be conserved. For sphere A and B, before collision: \[ m\bar{v}_{1} = m\bar{v}_{A}' + m\bar{v}_{B}', \]where \(\bar{v}_{A}'\) and \(\bar{v}_{B}'\) are the linear velocities of sphere A and sphere B immediately after the impact, respectively. Simplifying gives:\[\bar{v}_{1} = \bar{v}_{A}' + \bar{v}_{B}'. \]
03

Apply Coefficient of Restitution

Using the restitution relation:\[ \bar{v}_{B}' - \bar{v}_{A}' = -\bar{v}_{1}, \]as the relative velocity between them before collision was \(\bar{v}_{1}\).
04

Solve the System of Equations

We now have two equations: 1. \(\bar{v}_{1} = \bar{v}_{A}' + \bar{v}_{B}'\)2. \(\bar{v}_{B}' - \bar{v}_{A}' = -\bar{v}_{1}\)Solving these:- Adding them gives \( 2\bar{v}_{B}' = 0 \implies \bar{v}_{B}' = 0 \)- Substituting back gives \( \bar{v}_{A}' = \bar{v}_{1} \)
05

Determine Angular Velocities After Impact

Because there is no slipping, the angular velocity of sphere A, immediately after impact, remains the same as before the impact, i.e., \( \omega_{A}' = \frac{\bar{v}_{1}}{r} \). Sphere B starts rotating with \( \omega_{B}' = 0 \) initially, as it was at rest.
06

Consider Frictional Effects for Sphere A

Sphere A continues to experience kinetic friction until it rolls without slipping again. The friction decelerates it linearly and accelerates it angularly. Using \(f = \mu_k mg\) as the friction force:For linear deceleration:\[ a = \mu_k g \] Using \( v = v_0 + at \), and knowing sphere A starts with a velocity of \( v_1 \), friction acts to reduce this to \( v_r = r\omega \): \[ v_r = v_1 - \mu_k g t \]
07

Equate Linear and Angular Velocities for Rolling without Slipping

The angular acceleration due to friction is:\[ \alpha = \frac{\mu_k g}{r} \]Using \(\omega = \omega_0 + \alpha t \), where \( \omega_0 = \frac{v_1}{r} \) initially,\[ \omega_r = \frac{v_1}{r} + \frac{\mu_k g}{r} t. \]When rolling without slipping starts, \( v_r = r \omega_r \). Equating these:\[ v_1 - \mu_k g t = r \left( \frac{v_1}{r} + \frac{\mu_k g}{r} t \right). \]
08

Compute Final Velocities

Solving \( v_1 - \mu_k g t = v_1 + \mu_k g t \) yields:- \( v_r = \frac{v_1}{1 + \mu_k \frac{g}{r} t} \) and- \( \omega_r = \frac{v_r}{r} \).Since \( v_1 = v_r + \mu_k g t \), the velocity after rolling begins uniformly is:- Sphere A rolls at \((\bar{v}_A = \frac{2v_1}{1 + \mu_k \frac{g}{r} t}\)) with angular velocity \(\omega_A = \frac{\bar{v}_{A}}{r}\). Sphere B starts its kinetic friction motion not from rest and adjusts similarly.

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Key Concepts

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

Angular Velocity
The angular velocity of a rotating object is a measure of how fast it spins around a given axis. In the context of the exercise, when sphere A hits sphere B, both spheres gain or retain their angular velocities.
Before the collision, sphere A has an angular velocity, which can be calculated by the ratio of its velocity over its radius: \[ \omega_A = \frac{\bar{v}_1}{r} \].
This is because sphere A is rolling without slipping, so the point of contact with the ground is momentarily at rest.
After impact, sphere A maintains its angular velocity of \( \omega_A' = \frac{\bar{v}_1}{r} \) due to the lack of slipping at the moment of impact.
On the other hand, sphere B initially doesn't rotate. Hence, its angular velocity right after the impact remains zero until friction influences its motion.
The change in angular motion is a crucial part of analyzing elastic collisions with rolling objects.
Linear Momentum
Linear momentum is the product of an object's mass and its velocity (\( m \times v \)). In a closed system with no external forces, the total linear momentum is conserved during collisions.
In this exercise, when the moving sphere A collides with the stationary sphere B, the total linear momentum before the collision equals that after the collision: \[ m\bar{v}_{1} = m\bar{v}_{A}' + m\bar{v}_{B}' \].
This equation shows that the initial momentum of sphere A is shared between the two spheres post-collision.
Using this principle alongside the coefficient of restitution, we can construct equations to find the individual linear velocities of each sphere right after the impact.
Through solving these equations, we determine that sphere A continues with a linear velocity of \( \bar{v}_{1} \), and sphere B remains at rest immediately, highlighting the role of conservation in elastic collisions.
Rolling without Slipping
Rolling without slipping happens when an object rolls in such a way that its points of contact do not slide on the surface. This condition implies a relationship between linear velocity and angular velocity: \[ v = r \omega \].
For sphere A, right after the collision, it maintains its rolling without slipping condition, moving with the same linear and angular velocities it had before the impact.
For sphere B, however, this condition isn't initially satisfied right after impact since it started from rest and there's no angular motion introduced by the impact.
As time progresses, kinetic friction acts on sphere A and B to enforce this no-slipping condition once again. Sphere A will decelerate linearly due to friction, until its linear and angular velocities satisfy \( v_r = r \omega_r \), allowing it to roll smoothly without sliding over the surface.
Kinetic Friction
Kinetic friction plays a significant role in the dynamics post-collision when spheres return to rolling without slipping.
Kinetic friction acts directly opposite to the direction of motion. For sphere A, this results in a deceleration of its linear speed but an increase in angular speed due to the torque the frictional force provides.
The friction force can be represented as \( f = \mu_k mg \), where \( \mu_k \) is the coefficient of kinetic friction.
Over time, kinetic friction causes both spheres to eventually reach a state where they are rolling without slipping, thus ensuring \( v_r = r \omega_r \).
For sphere B, initially stationary, friction acts to bring it up to speed until its velocity and angular velocity align according to the rolling without slipping condition.
Understanding the influence of kinetic friction is crucial, as it governs the eventual movement dynamics of both spheres after the initial impact.

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