Suggestions:
- FBD
- Work-energy: For the spoked wheel at top (A), the mass is concentrated at the point in the center of the wheel (is a particle). Therefore, T = (1/2)*m*vG^2, where G is the center of the wheel. For the rim wheel at the bottom (B), the mass is concentrated at the outer surface (rigid body thin ring). Therefore, T = (1/2)*m*vG^2 + (1/2)*IG*omega^2. [Alternately, you can write, T = (1/2)*IC*omega^2 where C is the no-slip contact point.] The potential energy expression is the same for A as for B.
- Kinematics: No additional kinematics need for A. For B, use the no-slip condition at C.
- Solve
10 comments:
if T2 for part B can either equal (1/2)*m*vG^2 + (1/2)*IG*omega^2
or
(1/2)*IC*omega^2
Then can't we use the second one to solve for omega, then just plug it back into the first one to find vG instead of using kinematics to relate vB ot vC?
Also, for the moment of inertia equations in the back of the book, should this just be modeled as a circular rod?
We don't model the moment of inertia as a circular rod because we are told it has "negligible diameter."
I calculated my answer using Ic, then to double check i calculated it using Ig. Unfortunately, my answers were different by a factor of 2. I might be using the wrong Ig or Ic. For Ig I used m*r^2, and for Ic I used 1/2*m*r^2. Are these correct?
Also Adam, I believe you could use your method to solve for omega, but I think it's a lot more algebra than simply using kinematics. Using kinematic you just relate vc to vg, which is very easy since vc is 0, and math confirms instinct that omega=vg/r
to erik...the negligible diameter is oly for A, which we dont need the I for any way, but for B we do have a diameter to use for the moment of inertia.
also to janet...I am not totally sure, but i think we need to use the parallel axis theorem. that i beleive should fix our problem.
Remember that for part b you need to use mass moment inertia for a ring, not a disk!
Janet,
The way i did it, Ic=Ig+mr^2
Ic=2mr^2
You used Ic=.5mr^2...so I think that might be why your answers are differing by a factor of 2.
Steve-o,
Does the Mass moment of inertia of a ring = m*r^2 ?
Yes, m*r^2 is inertia for a thin ring.
Alright, question:
If you use T2=(1/2)(Ic)(omega^2), then can you say V2=mg(-xsin(theta))? If so, can you just set the T2s equal and say (1/2)m(Vg^2)+(1/2)Ig(omega^2)=(1/2)Ic(omega^2)? Then Vg=r*omega? That doesn't seem right, but how would you find V2 for point C?
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