Impact of rolling sphere with vertical rod

[sergiokapone]

Homework Statement

Uniform sphere of mass $m$ and radius $r$, rolling without slipping on a horizontal surface at a velocity $v_0$, hits the end of a rod of length $l$ and mass $10m$, hanging on a hinge (Figure). Determine how the ball will move and the rod after the impact. What time after, the motion of the ball goes into pure rolling? What is the speed of the roller? An impact is perfectly elastic and instantaneous. The coefficient of friction of the ball on the plane is $k$. Rolling friction neglected.

Homework Equations

Eqs of conservations:
Angular momentun: $mv_0l=\dfrac{10ml^2}{3}\omega-mvl$
Energy: $\underbrace{\dfrac{mv_0^2}{2}+\dfrac{2}{5}mr^2\dfrac{v_0^2}{2r^2}}_{\dfrac{7mv_0^2}{10}}=\dfrac{10ml^2}{3}\dfrac{\omega^2}{2}+\dfrac{mv^2}{2}$

The Attempt at a Solution

I feel that I made a conceptual error in the conservation laws, but I do not know where.

 

[sergiokapone]

Eqs of conservations:
Angular momentun: $mv_0l=\dfrac{10ml^2}{3}\omega-mvl$
Energy: $\underbrace{\dfrac{mv_0^2}{2}+\dfrac{2}{5}mr^2\dfrac{v_0^2}{2r^2}}_{\dfrac{7mv_0^2}{10}}=\dfrac{10ml^2}{3}\dfrac{\omega^2}{2}+\dfrac{mv^2}{2}$

I think I understand. The system of two equations is inconsistent. But if we assume that the sphere after being hit does not lose its spin moment, then everything works:
$\begin{equation} \left\{ \begin{array} \ mv_0l =\dfrac{10ml^2}{3}\omega &- mvl \\ \dfrac{mv_0^2}{2} =\dfrac{10ml^2}{3}\dfrac{\omega^2}{2} &+ \dfrac{mv^2}{2} \end{array} \right. \end{equation}\text{ without loss of spin angular momentum}$

Then, we have
$v=\dfrac{7}{13}v_0$, $\omega=\dfrac{6}{13} \dfrac{v_0}{l}$

But why the sphere does not stop spinning?

 

[ehild]

Homework Statement

Uniform sphere of mass $m$ and radius $r$, rolling without slipping on a horizontal surface at a velocity $v_0$, hits the end of a rod of length $l$ and mass $10m$, hanging on a hinge (Figure). Determine how the ball will move and the rod after the impact. What time after, the motion of the ball goes into pure rolling? What is the speed of the roller? An impact is perfectly elastic and instantaneous. The coefficient of friction of the ball on the plane is $k$. Rolling friction neglected.

Homework Equations

Eqs of conservations:
Angular momentun: $mv_0l=\dfrac{10ml^2}{3}\omega-mvl$
Energy: $\underbrace{\dfrac{mv_0^2}{2}+\dfrac{2}{5}mr^2\dfrac{v_0^2}{2r^2}}_{\dfrac{7mv_0^2}{10}}=\dfrac{10ml^2}{3}\dfrac{\omega^2}{2}+\dfrac{mv^2}{2}+?$

The Attempt at a Solution

I feel that I made a conceptual error in the conservation laws, but I do not know where.

You omitted the rotational energy of the ball from the energy equation.

ehild

 

[sergiokapone]

You omitted the rotational energy of the ball from the energy equation.

ehild

Yes, exactly! Thank you.

 

[ehild]

Then, we have
$v=\dfrac{7}{13}v_0$, $\omega=\dfrac{6}{13} \dfrac{v_0}{l}$

But why the sphere does not stop spinning?

The rod hits the ball at the height of its centre. So the rod does not exert torque about the centre of the ball. No torque, no change of angular momentum. (The force of friction is finite, it does not have effect on angular momentum in infinitesimally short time)

ehild

 

[sergiokapone]

The rod hits the ball at the height of its centre. So the rod does not exert torque about the centre of the ball. No torque, no change of angular momentum. (The force of friction is finite, it does not have effect on angular momentum in infinitesimally short time)

ehild

Oh, really! Thank you!
Now I start to calculate the time from bounce to pure roll.
I have two eqns:
$m\dfrac{dv}{dt}=-F_{f}$
$\dfrac{2}{5}mr^2\dfrac{d\omega}{dt}=-F_{f}r$
Here not mess with the signs?

 

[sergiokapone]

From integrating above eqns:
$\dfrac{2}{5}\omega r-v=const$
Constant determine from initial conditions, e.g. $const=\dfrac{2}{5}\dfrac{v_0}{r} r-\dfrac{7}{13}v_0=-\dfrac{9}{65}v_0$
When will the pure rolling $v=\omega r$, then $\dfrac{2}{5}\omega r-\omega r= -\dfrac{9}{65}v_0$
E.g.
$\omega r = \dfrac{3}{13}v_0$

$v=\dfrac{3}{13}v_0$ - velocity of pure rolling.
Then
$v_{at\; the\; finish\; of\; sliding,\; or\; starting\; pure\; roll}=v_{initial\; velocity\; after\; bounce} +at$
or
$\dfrac{3}{13}v_0-\dfrac{7}{13}v_0=at$
$-\dfrac{4}{13}v_0=at$
$t=-\dfrac{4}{13}v_0\dfrac{1}{a}$
From the eqn of center mass motion $a=-kg$

Answer $t=\dfrac{4}{13}\dfrac{v_0}{kg}$

That does not match with the answer book of problems $t=\dfrac{40}{91}\dfrac{v_0}{kg}$

 

[ehild]

Take care of the signs.
After collision, the ball travels to the right and spins anticlockwise. When it starts to roll it travels to the right with velocity v and rotates clockwise with ω=v/r. The friction accelerates clockwise rotation.

ehild

 

[sergiokapone]

Take care of the signs.
After collision, the ball travels to the right and spins anticlockwise. When it starts to roll it travels to the right with velocity v and rotates clockwise with ω=v/r. The friction accelerates clockwise rotation.

ehild

I was not thinking. Thanks!

Right calculations:

From integrating above eqns:
$\dfrac{2}{5}\omega r-v=const$
Constant determine from initial conditions, e.g. $const=\dfrac{2}{5}\dfrac{v_0}{r} r-\dfrac{7}{13}v_0=-\dfrac{9}{65}v_0$
When will the pure rolling $v=-\omega r$, then $\dfrac{2}{5}\omega r+\omega r= -\dfrac{9}{65}v_0$
E.g.
$\omega r = - \dfrac{9}{91}v_0$

$v=\dfrac{9}{91}v_0$ - velocity of pure rolling.
Then
$v_{at\; the\; finish\; of\; sliding,\; or\; starting\; pure\; roll}=v_{initial\; velocity\; after\; bounce} +at$
or
$\dfrac{9}{91}v_0-\dfrac{7}{13}v_0=at$
$-\dfrac{4}{13}v_0=at$
$t=-\dfrac{40}{91}v_0\dfrac{1}{a}$
From the eqn of center mass motion $a=-kg$

Answer $t=\dfrac{40}{91}\dfrac{v_0}{kg}$

 

[ehild]

Nice work!

ehild

 

[sergiokapone]

Nice work!

ehild

Thanks for the help!