Weight and Wheel (Linear and Angular Acceleration)

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SUMMARY

The discussion focuses on calculating the angular speed of a bicycle wheel when a block of mass m falls a distance h, with two scenarios based on the radius of attachment (ra and rb). The conservation of mechanical energy is applied, leading to the equations K=1/2mv^2 and U=mgh. The derived formula for angular speed w is w=SQRT((2mghr^2)/(mr^2+I))/r. A common point of confusion arises regarding the dependence of the solution on mass m and radius r, which was clarified through the correct use of notation.

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  • Understanding of mechanical energy conservation principles
  • Familiarity with moment of inertia (I) concepts
  • Knowledge of angular velocity and its relation to linear velocity (V=rw)
  • Basic algebraic manipulation skills for solving equations
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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:
 

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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!
 

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