My Math Forum Why we have to add I ω to final angular momentum; Lf?

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 July 5th, 2019, 04:41 AM #11 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 Thanks from topsquark and romsek
July 8th, 2019, 07:52 AM   #12
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Quote:
 Originally Posted by topsquark The clay should not make the wheel turn...The clay is coming in straight on. (To be more prescise: The moment vector $\displaystyle \vec{r}$ and the velocity $\displaystyle \vec{v}$ are along the same line and thus $\displaystyle \vec{v} \times \vec{r} = 0$, so no torque should be applied from the collision.) -Dan
If I kick football, It rolls. Here clay hits wheel so wheel rotates. Why clay should not make the wheel turn?

July 8th, 2019, 02:27 PM   #13
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Quote:
 Originally Posted by Ganesh Ujwal If I kick football, It rolls. Here clay hits wheel so wheel rotates. Why clay should not make the wheel turn?
It seems I might have misread the question. I had taken the wheel to be fixed in place. If it is allowed to move then yes, the wheel would start to rotate.

-Dan

July 8th, 2019, 11:00 PM   #14
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 Originally Posted by topsquark Because angular momenta (about the same axis of rotation) are additive. The wheel is spinning ( $\displaystyle I \omega$ ) and the clay is rotating about the same axis ( with an amout of angular momentum $\displaystyle mvr$ ).
How can you tell both are rotating about the same axis?

Why $\displaystyle I \omega$ for spinning wheel and $\displaystyle mvr$ for clay? Clay is not symmetrical, so we have to use $\displaystyle mvr$ for it?

Last edited by Ganesh Ujwal; July 8th, 2019 at 11:02 PM.

July 9th, 2019, 01:44 AM   #15
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 Originally Posted by Ganesh Ujwal How can you tell both are rotating about the same axis? Why $\displaystyle I \omega$ for spinning wheel and $\displaystyle mvr$ for clay? Clay is not symmetrical, so we have to use $\displaystyle mvr$ for it?
The clay sticks to the wheel, so it must be in rotational motion about the axis of the wheel. The mvr is the amount of angular momentum the clay has about this axis before as it hits the wheel.

Though I was wrong about the clay not rotating the wheel, as it would roll away after the impact, the clay will simply not have enough momentum to make the wheel rotate with that amount of angular speed. The clay would have to hit the wheel at a different position to do that. (Basically momentum before = momentum after, so mv = (M + m)V, where M is the mass of the wheel and V the linear speed of the wheel and clay after collision. Then use the rolling without slipping condition $\displaystyle v = r \omega$.)

-Dan

Addendum: For the clay, as stated in the problem, it really has not angular speed about the center of the wheel because it is coming in along the direction of a radius. (ie. the moment arm is 0). When it is stuck to the wheel we have $\displaystyle I_{clay} = mr^2$.

Last edited by topsquark; July 9th, 2019 at 01:47 AM.

July 9th, 2019, 06:03 AM   #16
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Quote:
 Originally Posted by topsquark Then use the rolling without slipping condition $\displaystyle v = r \omega$
Why I have to use rolling without slipping condition? What is this new thing:without slipping condition?

 July 9th, 2019, 02:50 PM #17 Global Moderator   Joined: Dec 2006 Posts: 20,965 Thanks: 2214 The problem states the wheel is set into pure rolling motion, which means no slipping. The wheel rotates about a moving axis, which is always its horizontal line of contact with the ground. Thanks from topsquark

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