- #1

Nexus99

- 103

- 9

- Homework Statement
- posted in solution

- Relevant Equations
- Conservation of angular momentum, kinetic energy etc.

A hollow rod closed at the ends A and B, has mass M and length R. The rod is free

to rotate on a horizontal frictionless plane around the z axis passing through A

and coming out of the sheet. A body can slide without friction inside the cavity

point mass m.

Initially the rod is stationary and the mass m is placed, also stationary, at a distance

negligible but finished by A. A ball of mass m0, with velocity ##\vec{v_0}## = (0,-v0, 0), elastically hits it in B. Calculate:

1. the speed of the ball ##\vec{v_1}## and the angular speed ##\vec{\omega_i}## I of the rod immediately after the impact;

2. the impulse of the constraint reaction ##\vec{J}##

After the impact, the mass m starts up and reaches the end B. Calculate:

3. the angular speed of the rod ##\vec{\omega_{f}}## and the module of the speed ##v_f## of the mass m

with respect to the rod when m reaches B;

4. assuming that the mass m bounces elastically in B, calculate the speed

angle ##\vec{\omega_{A}}## of the rod when m returns to A;

In the event that the rod is not constrained to A and the impact is completely inelastic determine:

5. the equation of motion of the overall center of mass of the ##\vec{r_{CM}}## system;

6. the angular velocity ##\vec{\omega_{CM}}## with respect to an axis passing through the center of mass.

1) angular momentum and kinetic energy is conserved:

## -m_0 v_0 R = m_0 v_1 R + I \omega_1 ##

## \frac{1}{2} m_0 v_0^2 = \frac{1}{2} I \omega_{i}^2 + \frac{1}{2} m_0 v_1^2 ##

Solving the system i got:

## v_1 = \frac{(I-R^2 m_0) v_0}{I + R^2 m_0}##

## \omega_i = -\frac{2 R m_0 v_0}{I + R^2 m_0}##

where ## I = \frac{M R^2}{3} ##

## \vec{v_1} = (0,v_1,0)## and ##\vec{\omega_i} = (0,0,\omega_{i}) ##

2) Impulse of the costraint reaction = - Impulse generated by the ball

## J = -(m_0 v_1 - m_0 v_0) = \frac{2R^2 m_0^2 v_0}{I + R^2 m_0} ##

##\vec{J} = (0,J,0) ##

3) Angular momentum and kinetic energy is conserved

## I \omega_{i} = I \omega_{f} + m v_{f} R##

## \frac{1}{2} I \omega_i^2 = \frac{1}{2} I \omega_f^2 + \frac{1}{2} m v_f^2 ##

and i got

##v_f = \frac{2 I R \omega_i}{I + R^2m_0} ##

##\omega_f = \frac{\omega_{i}(I - R^2 m_0 }{I + R^2m_0} ##

##\vec{\omega_{f}} = (0,0, \omega_{f}) ##

4) momentum and kinetic energy is conserved

## - m v_f = m v_A ##

## \frac{1}{2} I \omega_f^2 + \frac{1}{2} m v_f^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} I \omega_A^2##

and i got

##\omega_f = \omega_A##

##\vec{\omega_A} = (0,0, - \omega_f )##

5) i don't know, i think that ##x_{CM}## is constant since total ##p_{x}## is constant and equal to 0

So i would say ##x_{CM} = \frac{R(2m_0 + M)}{2(m + M + m_0)}##

Neither the momentum along the y-axis nor the kinetic energy is conserved, so i think that the CM along y-axis is moving uniform linear motion but i don't know how to find ##v_{y_{CM}}##

Can you help me?

to rotate on a horizontal frictionless plane around the z axis passing through A

and coming out of the sheet. A body can slide without friction inside the cavity

point mass m.

Initially the rod is stationary and the mass m is placed, also stationary, at a distance

negligible but finished by A. A ball of mass m0, with velocity ##\vec{v_0}## = (0,-v0, 0), elastically hits it in B. Calculate:

1. the speed of the ball ##\vec{v_1}## and the angular speed ##\vec{\omega_i}## I of the rod immediately after the impact;

2. the impulse of the constraint reaction ##\vec{J}##

After the impact, the mass m starts up and reaches the end B. Calculate:

3. the angular speed of the rod ##\vec{\omega_{f}}## and the module of the speed ##v_f## of the mass m

with respect to the rod when m reaches B;

4. assuming that the mass m bounces elastically in B, calculate the speed

angle ##\vec{\omega_{A}}## of the rod when m returns to A;

In the event that the rod is not constrained to A and the impact is completely inelastic determine:

5. the equation of motion of the overall center of mass of the ##\vec{r_{CM}}## system;

6. the angular velocity ##\vec{\omega_{CM}}## with respect to an axis passing through the center of mass.

1) angular momentum and kinetic energy is conserved:

## -m_0 v_0 R = m_0 v_1 R + I \omega_1 ##

## \frac{1}{2} m_0 v_0^2 = \frac{1}{2} I \omega_{i}^2 + \frac{1}{2} m_0 v_1^2 ##

Solving the system i got:

## v_1 = \frac{(I-R^2 m_0) v_0}{I + R^2 m_0}##

## \omega_i = -\frac{2 R m_0 v_0}{I + R^2 m_0}##

where ## I = \frac{M R^2}{3} ##

## \vec{v_1} = (0,v_1,0)## and ##\vec{\omega_i} = (0,0,\omega_{i}) ##

2) Impulse of the costraint reaction = - Impulse generated by the ball

## J = -(m_0 v_1 - m_0 v_0) = \frac{2R^2 m_0^2 v_0}{I + R^2 m_0} ##

##\vec{J} = (0,J,0) ##

3) Angular momentum and kinetic energy is conserved

## I \omega_{i} = I \omega_{f} + m v_{f} R##

## \frac{1}{2} I \omega_i^2 = \frac{1}{2} I \omega_f^2 + \frac{1}{2} m v_f^2 ##

and i got

##v_f = \frac{2 I R \omega_i}{I + R^2m_0} ##

##\omega_f = \frac{\omega_{i}(I - R^2 m_0 }{I + R^2m_0} ##

##\vec{\omega_{f}} = (0,0, \omega_{f}) ##

4) momentum and kinetic energy is conserved

## - m v_f = m v_A ##

## \frac{1}{2} I \omega_f^2 + \frac{1}{2} m v_f^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} I \omega_A^2##

and i got

##\omega_f = \omega_A##

##\vec{\omega_A} = (0,0, - \omega_f )##

5) i don't know, i think that ##x_{CM}## is constant since total ##p_{x}## is constant and equal to 0

So i would say ##x_{CM} = \frac{R(2m_0 + M)}{2(m + M + m_0)}##

Neither the momentum along the y-axis nor the kinetic energy is conserved, so i think that the CM along y-axis is moving uniform linear motion but i don't know how to find ##v_{y_{CM}}##

Can you help me?

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