
#1
Jun1204, 03:51 PM

P: 1,370

A person stadns on a platform, initially at rest, that can rotate freely without friction. The moment of inertia of the person plus the platform is [itex]I_p[/itex]. The person holds a spinning bicycle wheel with axis horizontal. The wheel has momemnt of inertia [itex]I_w[/itex] and angular velocity [itex]\omega_w[/itex]. What will the angular velocity [itex]\omega_p[/itex] of the platform if the person moves the axis of the wheel so that it points vertically upward.
Using conservation of momentum: [tex]\vec{L}_w = \vec{L'}_w + \vec{L}_p[/tex]where [tex]\begin{align*}Solving for [itex]\omega_1[/itex] and [itex]\omega_2[/itex] gives [tex]\begin{align*}And since [itex]\omega_p = \sqrt{\omega_1^2 + \omega_2^2}[/itex] then [tex]\omega_p = \frac{\sqrt{2}I_w\omega_w}{I_p}[/tex]According to my book though, [tex]\omega_p = \frac{I_w\omega_w}{I_p}[/tex]Is it just me, or is my book on crack? 



#2
Jun1204, 07:10 PM

P: 1,370





#3
Jun1204, 07:12 PM

Mentor
P: 40,907

The book seems correct to me. Note that angular momentum is only conserved about the vertical axis: the platform is constrained and only freely rotates about that vertical axis.
So: initial angular momentum about vertical axis is zero! The wheel gains [itex]+ I_w\omega_w[/itex], so the platform must gain [itex] I_w\omega_w[/itex] 



#4
Jun1204, 07:18 PM

P: 1,047

Spinning Bicyle Wheel, Person, and Platform
The book's answer looks OK to me.
Initially, the angular momentum about the vertical axis is 0, so conservation of momentum requires that this remains true. When the wheel's axis turns vertical, the wheel now has angular momentum about the vertical axis equal in magnitude to it's initial angular momentum, L_{w} = I_{w}ω_{w}. The (person+wheel) must have angular momentum of equal magnitude in the opposite direction in order for the net total vertical angular momentum to remain 0. Therefore, [tex]I_p\omega_p = I_w\omega_w[/tex] [tex]\omega_p = \frac{I_w\omega_w}{I_p}[/tex] 



#5
Jun1204, 07:20 PM

P: 1,047

Yeah, like he said...




#6
Jun1204, 08:33 PM

P: 1,370

What!? I didn't know angular momentum was picky about the axis it chooses to remain conservative in. Now I'm really confused. So how do I determine the axis where angular momentum is conserved in general when I have a system of rotating objects?




#7
Jun1204, 10:54 PM

P: 1,047

Momentum, whether linear or angular, is a vector so conservation requires that you consider the direction as well as the magnitude.
As far as which axis...that's determined by the physical characteristics of the system you're looking at. As the wheel ends up spinning on a vertical axis, and the man is on a platform that spins freely on a vertical axis, the vertical is the axis you have to worry about. The next question is, what happened to the momentum about the horizontal axis? Why does that seem not to have been conserved? If you think a bit about that aspect, you should come up with the answer. 



#8
Jun1204, 11:44 PM

P: 1,370





#9
Jun1304, 07:09 AM

Mentor
P: 40,907

Of course, if you include the earth as part of your systemthen angular momentum is conserved in all directions! 



#10
Jun1304, 09:11 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Just adding a few comments to what Doc Al and gnome have said:
Consider the system consisting of platform, person and wheel. The platform has an axle through it (fixed to the ground), about which it may rotate freely. (The "axle" also has a "plate" attached to it, which the platform rests upon; there is no friction between the platform or the "plate" either.) Hence, no external forces exist which may produce torques about the axis (These would have been frictional forces between platform and axle) Therefore, in the interval (0,T) between initial state and final state, the momentof momentum equation, with the torques computed relative to the C.M of the system, we have: [tex]\vec{\tau}=\frac{d\vec{S}_{C.M}}{dt},\vec{\tau}\cdot\vec{k}=0[/tex] Here, [tex]\vec{k}[/tex] is parallell to the axle, [tex]\vec{S}_{C.M}[/tex] is the angular momentum of the system with respect to C.M, while [tex]\vec{\tau}[/tex] is the net torque of external forces acting on the system. Clearly, we therefore have: [tex]\vec{S}_{C.M}(0)\cdot\vec{k}=\vec{S}_{C.M}(T)\cdot\vec{k}[/tex] This gives the relation in the book by identifying [tex]\vec{S}_{C.M}(0)\cdot\vec{k}=0[/tex] 



#11
Jun1304, 10:37 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Note:
The external, torqueproducing forces are normal forces from the axle onto the platform . In reality, this will cause the platform to wobble slightly between initial state and final state. The force responsible for the platform's rotation, is the frictional force between the person's feet and the platform. 


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