July 3rd, 2019, 04:59 AM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 406 Thanks: 1  Why we have to add I ω to final angular momentum; Lf?
A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1m. Assuming that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately. Sol: $\displaystyle L_{i} = mvr = 1*10*1 = 10$ $\displaystyle L_{f} = mv_{cm}r + Iω$ $\displaystyle = 20*rω*r + \large\frac{mr^2 ω}{2}\normalsize = 20ω + 10ω$ $\displaystyle L_{f} =30 ω$ $\displaystyle L_{i} = L_{f}$ $\displaystyle 10 = 30 ω$ $\displaystyle ω = \large\frac {1}{3} \;\normalsize rad/sec$ Why do we have to add $\displaystyle Iω$ to final angular momentum; $\displaystyle L_{f}$ ? Last edited by skipjack; July 3rd, 2019 at 08:13 AM. 
July 3rd, 2019, 05:50 AM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff. 
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 $\displaystyle mvr$ ). That's the answer you are looking for, but frankly I don't like this problem. 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.) And even if we take the spirit of the problem and say that it does make the wheel rotate anyway, it would not spin with anything like the angular speed that the problem is saying it does. Dan Last edited by topsquark; July 3rd, 2019 at 05:54 AM. 
July 3rd, 2019, 05:58 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573 
Something is wrong with the problem or your set up ... Given your sketch, the initial angular momentum of of the clay mass w/respect to the center of the wheel is zero because the clay mass strikes the wheel along an axis through its center. In this case, the wheel will roll only if the friction of the surface it rests on generates a sufficient torque to make it roll. Are you sure the clay did not strike the wheel offset from its center? 
July 3rd, 2019, 06:55 AM  #4 
Member Joined: Jun 2019 From: USA Posts: 69 Thanks: 27 
If the wheel has pure rolling motion, there clearly must be nontrivial friction with the surface (a reasonable physical assumption, anyway). If there is friction, the instant the clay hits the wheel, the impact force produces a moment about the point of contact with the surface. Or, the friction force produces a moment about the centre of mass of the wheel (or wheelclay system). From any point of view, it should cause the wheel to spin in the direction of rolling forward. If you need more convincing, go watch some billiards. Edit: Never mind, I didn't look at the solution closely enough. If the friction imparted angular momentum to the system, then clearly you couldn't use conservation of momentum (angular or linear). I don't understand that $\displaystyle L_i$ term, either. Are they claiming the clay hits the very top of the wheel and sticks? Edit 2: Clearly, I can't read, either. I see now Skeeter said the same thing I did. Last edited by DarnItJimImAnEngineer; July 3rd, 2019 at 07:15 AM. 
July 3rd, 2019, 09:50 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160 
The axis of rotation of the wheel is its line of contact with the ground.

July 3rd, 2019, 03:01 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573  
July 3rd, 2019, 03:42 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 
I think the issue is that the problem isn't (and isn't trying to be) strictly realistic. They simply state that there's friction enough to cause pure rolling motion and wave their hands about actual friction coefficients, and moments about an axis etc. They simply want the linear momentum of the clay converted into angular momentum of the wheel. 
July 3rd, 2019, 05:16 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160  
July 3rd, 2019, 11:59 PM  #9 
Senior Member Joined: Apr 2014 From: UK Posts: 936 Thanks: 335  
July 4th, 2019, 08:25 PM  #10 
Member Joined: Jun 2019 From: USA Posts: 69 Thanks: 27 
If we take point C to be the point on the wheel that contacts the ground while it is stationary. Since the friction occurs at this point during impact, then the frictional moment about C is zero, and we can ignore it in terms of $\displaystyle \vec{\tau}={d\vec{L}}/{dt}$. If the clay hits in line with the centre of the wheel, then the instantaneous momentum about C would be $\displaystyle 10 kg\cdot m^2/s$. It looks like two problems, though. 1) The rotational inertia of the wheel about C is not $\displaystyle mr^2/2$. You have to use the parallel axis theorem. 2) If point C is attached to the wheel, then any coordinate system with C as the origin will be noninertial, and thus you can't apply conservation of momentum normally. It's been a loooong time since solid dynamics, so someone correct me if I'm wrong, but I don't see a way to predict this without making assumptions about either the friction coefficient(s) or the duration of impact. That doesn't make physical or dimensional sense, though. You can't "convert" linear momentum to angular without a radius of rotation (which is zero for the incoming clay). 

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