Weight and Wheel (Linear and Angular Acceleration)

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The discussion revolves around calculating the angular speed of a bicycle wheel when a mass is attached and allowed to fall. For part (a), the user derives the angular speed based on energy conservation principles, leading to a formula that incorporates the moment of inertia and radius. In part (b), they question whether the approach remains the same with a different radius, expressing confusion about the dependence on mass and radius. Ultimately, the user resolves their confusion by realizing a notation error with the angular velocity symbol. The thread highlights the complexities of applying physics concepts to rotational motion problems.
tizzful
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Weight and Wheel ! (Linear and Angular Acceleration)

Homework Statement


Consider a bicycle wheel that initially is not rotating. A block of mass is attached to the wheel and is allowed to fall a distance . Assume that the wheel has a moment of inertia, I, about its rotation axis.
A)Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius ,ra. Find the angular speed of the wheel after the block has fallen a distance h, for this case.
b)Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius ,rb. Find the angular speed of the wheel after the block has fallen a distance h, for this case.

Homework Equations



K=1/2mv^2 U=mgh Mechanical energy conserved therefore E=K+U
V=rw
Images attached

The Attempt at a Solution


(a)U1+K1=U2+K2
mgh+0=0+1/2mv^2+1/2Iw^2
2mgh=mv^2+Iw^2
2mgh=mv^2+I(V/r)^2
(2mghr^2)/(mr^2+I)=v^2
V=SQRT((2mghr^2)/(mr^2+I))
w=v/r
w=SQRT((2mghr^2)/(mr^2+I))/r

(b) the same? just changing the r values?

I entered that as my answer but it says that it does not depend on the variable m and r but the only way to get rid of them is if the moment of intertia wasn't I.. So now I'm very confused and have no way of figuring out how to cancel them...
Help?
Thank you!:smile:
 

Attachments

  • MRB_ke_a.jpg
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  • MRB_ke_b.jpg
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tizzful said:

Homework Equations



K=1/2mv^2 U=mgh Mechanical energy conserved therefore E=K+U
V=rw
Images attached

The Attempt at a Solution


(a)U1+K1=U2+K2
mgh+0=0+1/2mv^2+1/2Iw^2
2mgh=mv^2+Iw^2
2mgh=mv^2+I(V/r)^2
(2mghr^2)/(mr^2+I)=v^2
V=SQRT((2mghr^2)/(mr^2+I))
w=v/r
w=SQRT((2mghr^2)/(mr^2+I))/r
Looks fine to me, but please simplify by canceling that outside r.

(b) the same? just changing the r values?

I entered that as my answer but it says that it does not depend on the variable m and r but the only way to get rid of them is if the moment of intertia wasn't I..
That makes no sense to me. (I assume you've stated the problem completely and that there's no additional information given.)
 
Oh thank you, it just turned out I wasn't using the greek w and so basically got the question wrong! ugh very annoying!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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