Challenge problem, week of 11/29/10, Part II

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Sequel to sliding puck problem. If you haven't yet figured out part I, that would be a good place to start. I've also posted another challenge in case people are tired of the sphere.

What if, instead of a sliding point-mass puck, we roll an object down the sphere? Object rolls without slipping and has given moment of inertia.

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I used the same method as for part I but with Kinetic Energy having both rotational and linear components. I have begun the derivation using some geometric observations made in Part I of the problem. *note: this derivation will be hard to follow without first walking through the solution to Part I of the challenge problem.

Finding the changed velocity: (This time around, I figured the kinetic energy at the time the ball leaves the basketball would have both rotational and linear kinetic energy because it is rolling)\

Eventually I changed omega, rotational velocity, into a linear expression to solve for velocity.

$\Delta U = \Delta K \Rightarrow 1/2mv^2 + 1/2I\omega ^2 \Rightarrow mgh = 1/2mv^2 +1/2I(v^2/r^2) \Rightarrow 2mgh = (m+I/r^2)v^2 \Rightarrow 2mghr^2 = (I+mr^2)v^2 \Rightarrow v^2 =\frac{2mghr^2}{I+mr^2}$

Just like part I, substitute the velocity squared expression into the cosine expression.

$cos\theta = v^2/r = (\frac{2mghr^2}{I+mr^2})/rg = \frac{2mhr}{I+mr^2}$

recalling the alternate expression from Part I, and recalling that the vertical height "h" is the vertical height "r"

$\frac{2mhr}{I+mr^2} = \frac{r-k}{r} \Rightarrow \frac{2mkr}{I+mr^2} = \frac{r-k}{r} \Rightarrow \frac{2mkr^2}{I+mr^2} = r-k$

$2mkr^2 = (r-k)(I+mr^2) \Rightarrow 2mkr^2 = Ir -Ik +mr^3 -mkr^2 \Rightarrow 2mkr^2 +Ik +mkr^2= Ir +mr^3 \Rightarrow k(2mr^2+I+mr^2) = Ir+mr^3 \Rightarrow k= \frac{Ir+mr^3}{I +3mr^2}$

After isolating the "k" terms, I had found a solution.

Edited by willorn
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• 4 weeks later...

Willorn,

Good job with the energy conservation treatment. Looks like you're confusing the radius of the rolling object with the radius of the basketball (generally, they're not the same). In the first two lines, r can only correctly stand for the radius of the rolling object. On your third line, you pull out $\cos \theta= \frac{v^2}{r g}$, where r now represents the radius of the big sphere. You then let the two r mingle, which is generally not a good thing.

The good part: Dimensionally, your answer is good. It even reduces to R/3 if we take I to zero. Looks like you found a valid answer in the special case where r = R (maybe we could use R for the basketball and r for the little object?

Edited by Anthony Frachioni
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A quick re-work of my work yeilded a much stranger looking answer, but since I could find no fault in my original method, I can only assume my new answer to be the simplest form solution. (Sidenote: the "IK" term towards the end should be positive after it is added to both sides, I got caught up in Latex)

$cos\theta = \frac{v^2}{Rg} = \frac {R-K}{R} \Rightarrow \frac{\frac{2mgkr^2}{I+mr^2}}{Rg} = \frac {R-K}{R} \Rightarrow \frac {2mkr^2}{I+mr^2} = R-K \Rightarrow 2mkr^2 = (R-K)(I+mr^2) \Rightarrow 2mkr^2 = IR -IK +mRr^2-mkr^2 \Rightarrow 3mkr^2 - IK = IR +mRr^2 \Rightarrow k(3mr^2 -I)= IR+mRr^2$

$k = \frac {IR +mRr^2}{3mr^2+I}$

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