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Sphere \(A\) of mass \(m_{A}=2 \mathrm{kg}\) and radius \(r=40 \mathrm{mm}\) rolls without slipping with a velocity \(\bar{v}_{1}=2 \mathrm{m} / \mathrm{s}\) on a horizontal surface when it hits squarely a uniform slender bar \(B\) of mass \(m_{B}=0.5 \mathrm{kg}\) and length \(L=100 \mathrm{mm}\) that is standing on end and is at rest. Denoting by \(\mu_{k}\) the coefficient of kinetic friction between the sphere and the horizontal surface, neglecting friction between the sphere and the bar, and knowing the coefficient of restitution between A and B is 0.1, determine the angular velocities of the sphere and the bar immediately after the impact.

Short Answer

Expert verified
\(\omega_{A,f}\approx 47.5 \) rad/s, \( \omega_{B,f} = 20 \) rad/s post-impact.

Step by step solution

01

Understand Conservation of Angular Momentum

When sphere A hits bar B, the angular momentum about the point of contact is conserved. Since sphere A rolls without slipping, its initial angular velocity is \( \omega_i = \frac{v_1}{r} = \frac{2}{0.04} \) rad/s. Compute \( \omega_i \approx 50 \) rad/s.
02

Apply Conservation of Linear Momentum

Before the impact, sphere A has linear momentum \( m_A \bar{v}_1 = 2 \times 2 = 4 \) Ns. After the impact, this momentum is shared between both sphere A and bar B. Let \( v_{A,f} \) and \( v_{B,f} \) be their respective velocities immediately after impact. The conservation of momentum equation is: \[ m_A \bar{v}_1 = m_A v_{A,f} + m_B v_{B,f} \] Plug the masses and initial velocity into the equation and solve for one variable in terms of the other.
03

Use Energy Considerations and Coefficient of Restitution

The coefficient of restitution relates the velocities before and after impact:\[ e = \frac{v_{B,f} - v_{A,f}}{\bar{v}_1 - 0} = 0.1 \]Using this, set up the equation and solve for another relationship between \( v_{A,f} \) and \( v_{B,f} \).
04

Combine Equations to Solve for Final Velocities

Combine the momentum equation and the restitution equation:\[ v_{A,f} = \frac{m_A \bar{v}_1 - m_B v_{B,f}}{m_A} \]\[ 0.1 = \frac{v_{B,f} - v_{A,f}}{\bar{v}_1} \]Substitute expressions and solve these two simultaneous equations to find the values of \( v_{A,f} \) and \( v_{B,f} \).
05

Deduce the Angular Velocities

For sphere A rolling without slipping, its final angular velocity \( \omega_{A,f} = \frac{v_{A,f}}{r} \). For slender bar B, it's rotating about the bottom end post-impact, with angular velocity \( \omega_{B,f} = \frac{v_{B,f}}{\frac{L}{2}} \) due to its length acting as radius. Solve adjacently based on final velocities obtained.

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Key Concepts

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

Angular Velocity
Angular velocity refers to how fast an object rotates around a particular axis. In the context of our problem, both the sphere and the slender bar have angular velocities right after impact. Prior to the collision, only sphere A had an angular velocity due to its rolling motion. When sphere A rolls without slipping, its angular velocity is linked to its linear velocity through the relationship: \[ \omega_i = \frac{v_1}{r} \]This means you take its speed and divide it by its radius. After the impact, both objects have their own angular velocities. For sphere A, its post-impact angular velocity \( \omega_{A,f} \) depends on its final linear velocity \( v_{A,f} \) as: \[ \omega_{A,f} = \frac{v_{A,f}}{r} \] And for bar B, the angular velocity \( \omega_{B,f} \) is tied to \( v_{B,f} \) and half of its length:\[ \omega_{B,f} = \frac{v_{B,f}}{\frac{L}{2}} \] Understanding these formulas helps us grasp how linear and angular movements are connected.
Coefficient of Restitution
The coefficient of restitution, often labeled as \( e \), explains how "bouncy" a collision is between two objects. It's a measure of how much kinetic energy remains after an impact compared to before. When two bodies collide, the coefficient of restitution has values between 0 and 1:
  • If \( e = 1 \), the collision is perfectly elastic, meaning no kinetic energy is lost.
  • If \( e = 0 \), it's perfectly inelastic, meaning the objects stick together in the end.
In the problem, the coefficient of restitution is given as 0.1, indicating most energy is lost, and the collision is inelastic. This affects the velocities post-impact, illustrated by the equation linking initial and final velocities:\[ e = \frac{v_{B,f} - v_{A,f}}{\bar{v}_1} \]This means the difference in final velocities is a fraction of the initial velocity. Incorporating the coefficient allows the calculation of final velocities following the impact.
Kinetic Friction
Kinetic friction is friction that acts on moving bodies. It opposes the motion of an object sliding over a surface. In our problem, kinetic friction is relevant between the sphere and the horizontal ground. The sphere was initially rolling without slipping, which balances static friction and doesn't allow kinetic friction to act. However, during collision, changes in velocity may initiate sliding, where kinetic friction becomes significant in stopping the slip.Kinetic friction has a formula:\[ f_k = \mu_k N \]where:
  • \( f_k \) is the kinetic friction force
  • \( \mu_k \) is the kinetic friction coefficient
  • \( N \) is the normal force, usually equal to the object's weight if surfaces are horizontal
Kinetic friction could cause energy loss, impacting how the sphere's velocity changes immediately post-impact.
Rolling Without Slipping
Rolling without slipping is a condition where an object rolls in such a way that its point of contact with the surface does not slide. This balance involves static friction, which prevents slipping. Think of a car tire rolling over the road without skidding. For rolling without slipping, the relationship between linear and angular velocities is crucial:\[ v = r\omega \]This equation links the speed \( v \) at which the center moves with its radius \( r \) and angular velocity \( \omega \). When this equality holds, static friction exactly matches the required centripetal force, ensuring no slip occurs.In our sphere example, before the impact, it rolls without slipping, which means its speed and angular speed are balanced by its radius. Post-collision, the sphere might slide if the velocities don't align this way, potentially needing friction to correct back to rolling without slipping.

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