- #1

giodude

- 30

- 1

- Homework Statement
- A rigid rod of length ##d## and mass ##m## is lying on a horizontal frictionless table and pivoted at the point ##P## on one end (shown in figure below). A point-like object of the same mass ##m## is moving to the right (see figure below). with speed ##v_{i}##. It collides elastically with the rod at the midpoint of the rod and rebounds backwards with speed ##v_{f}##. After the collision, the rod rotates clockwise about its pivot point ##P## with angular speed ##\omega_{f}##. The moment of inertia of a rod about the center of mass is ##I_{cm} = \frac{1}{12}md^{2}##.

Find the angular speed ##\omega_{f}##. Express your answer in terms of ##d##, ##m##, and ##v_{i}## as needed.

- Relevant Equations
- Elastic collision: ##\Delta K = 0##

Parallel axis theorem: ##I_{P} = I_{cm} + md^{2}##

Angular kinetic energy: ##K = \frac{1}{2}I_{P}^{sys} \omega_{f}^{2}##

Given that we're working with an elastic collision we want to populate the following system:

##k_{i} = k_{f}##

##p_{i} = p_{f}##

Solve for kinetic energy just before and after the collision:

##k_{i} = \frac{1}{2}mv_{i}^{2}##

##k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}I_{P}^{sys} \omega_{f}^{2}##

Solve for ##I_{P}## using parallel axis theorem:

##I_{P} = I_{cm} + md^{2} = \frac{1}{12}md^{2} + m(\frac{1}{2}d)^{2} = \frac{1}{12}md^{2} + \frac{1}{4}md^{2}##

##\frac{1}{12}md^{2} + \frac{1}{4}md^{2} = \frac{1}{3}md^{2}##

Plug moment of inertia about ##P##, ##I_{P}## in to our equation for final kinetic energy:

##k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}\frac{1}{3}md^{2} \omega_{f}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}md^{2} \omega_{f}^{2}##

Simplify ##k_{i} = k_{f}##:

##\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}md^{2} \omega_{f}^{2}##

##v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

Find ##p_{i}, p_{f}##:

##p_{i} = mv_{i}##

##p_{f} = mv_{f} + mv_{cm}##

The rod has a pivot P, so upon collision it's center of mass will have angular speed such that ##v_{cm} = \frac{d}{2}\omega_{f}##:

##p_{i} = mv_{i}##

##p_{f} = mv_{f} + m\frac{d}{2}\omega_{f}##

Solve the system:

##mv_{i} = mv_{f} + m\frac{d}{2}\omega_{f}##

##v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

##v_{f} = v_{i} - \frac{d}{2}w_{f}##

##v_{i}^{2} = (v_{i} - \frac{d}{2}\omega_{f})^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

##v_{i}^{2} = v_{i}^{2} - dv_{i}\omega_{f} + \frac{1}{4}d^{2} \omega_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

##dv_{i}\omega_{f} = \frac{7}{12}d^{2}\omega_{f}^{2}##

##\omega_{f} = \frac{12v_{i}}{7d}##

Hello, I know elastic collisions have ##\Delta K = 0## which is why it was used. However, I'm curious when we're allowed to use the coefficient of restitution, ##e##, to set the ratio ##\frac{v_{f}}{v_{i}} = 1##. I learned in textbook reading that this is usually a feature of elastic collisions however it doesn't seem to be the case here. So I'm curious (a) if this solution is correct and (b) when do I know what scenarios allow for the use of ##e = \frac{v_{f}}{v_{i}} = 1##. Thank you in advance!

##k_{i} = k_{f}##

##p_{i} = p_{f}##

Solve for kinetic energy just before and after the collision:

##k_{i} = \frac{1}{2}mv_{i}^{2}##

##k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}I_{P}^{sys} \omega_{f}^{2}##

Solve for ##I_{P}## using parallel axis theorem:

##I_{P} = I_{cm} + md^{2} = \frac{1}{12}md^{2} + m(\frac{1}{2}d)^{2} = \frac{1}{12}md^{2} + \frac{1}{4}md^{2}##

##\frac{1}{12}md^{2} + \frac{1}{4}md^{2} = \frac{1}{3}md^{2}##

Plug moment of inertia about ##P##, ##I_{P}## in to our equation for final kinetic energy:

##k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}\frac{1}{3}md^{2} \omega_{f}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}md^{2} \omega_{f}^{2}##

Simplify ##k_{i} = k_{f}##:

##\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}md^{2} \omega_{f}^{2}##

##v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

Find ##p_{i}, p_{f}##:

##p_{i} = mv_{i}##

##p_{f} = mv_{f} + mv_{cm}##

The rod has a pivot P, so upon collision it's center of mass will have angular speed such that ##v_{cm} = \frac{d}{2}\omega_{f}##:

##p_{i} = mv_{i}##

##p_{f} = mv_{f} + m\frac{d}{2}\omega_{f}##

Solve the system:

##mv_{i} = mv_{f} + m\frac{d}{2}\omega_{f}##

##v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

##v_{f} = v_{i} - \frac{d}{2}w_{f}##

##v_{i}^{2} = (v_{i} - \frac{d}{2}\omega_{f})^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

##v_{i}^{2} = v_{i}^{2} - dv_{i}\omega_{f} + \frac{1}{4}d^{2} \omega_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}##

##dv_{i}\omega_{f} = \frac{7}{12}d^{2}\omega_{f}^{2}##

##\omega_{f} = \frac{12v_{i}}{7d}##

Hello, I know elastic collisions have ##\Delta K = 0## which is why it was used. However, I'm curious when we're allowed to use the coefficient of restitution, ##e##, to set the ratio ##\frac{v_{f}}{v_{i}} = 1##. I learned in textbook reading that this is usually a feature of elastic collisions however it doesn't seem to be the case here. So I'm curious (a) if this solution is correct and (b) when do I know what scenarios allow for the use of ##e = \frac{v_{f}}{v_{i}} = 1##. Thank you in advance!