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STEP III 2007 question 11 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 11 solution

(i) Let O be the point of contact with the table edge, and H be the position of the hub. Define \theta to be the angle OH makes with the vertical.

To show the frictional force is zero, I think you just need to take moments about H - as the gravitational force and the reaction both act through H, any frictional force will result in a net torque being applied. But since the hub has zero moment of inertia any such torque will result in infinite angular acceleration, which is physically impossible.

For the next bit, conservation of energy gives us \frac{1}{2}a^2\dot{\theta}^2 = \frac{1}{2}u^2+ga(1-\cos\theta) and then differentiating gives us \ddot{\theta} = \frac{g}{a} \sin \theta

Now consider the y-coordinate of the hub:

y = a \cos \theta. Differentiate (twice):

\dot{y} = -a \dot{\theta} \sin \theta

\ddot{y} = - a\dot{\theta}^2\cos \theta -a \ddot{\theta}\sin\theta. Substitute in our values for \dot{\theta}^2, \ddot{\theta} to get:

a\ddot{y} = (-u^2+2ga(\cos \theta - 1)) \cos \theta - a^2 \frac{g}{a} \sin^2 \theta

=-(u^2+2ga)\cos \theta + 2ga \cos^2\theta -ga \sin^2 \theta

=-(u^2+2ga)\cos \theta + 3ga \cos^2\theta-ga

Now if there was no reaction force from the table, we'd have \ddot{y} = -g (full acceleration due to gravity). So at that point we have:

-ga = -(u^2+2ga)\cos \theta + 3ga \cos^2\theta-ga

0= \cos \theta((u^2+2ga)-3ga \cos \theta)

So \cos \theta = 0 or \cos \theta = \frac{u^2+2ga}{3ga}. The 2nd condition will occur before we get to \cos \theta = 0, so the wheel loses contact when \cos \theta = \frac{u^2+2ga}{3ga}. At this point, it has dropped a distance a(1-\cos \theta) = \frac{3ga - (u^2+2ga)}{3g} = \frac{ag-u^2}{3g} as desired.

(ii) I don't think any significant calculations are actually needed (or expected?) here. Again, take moments about the center of the wheel. The only torque is from the frictional force, but in this case the wheel has a non-zero moment-of-inertia, so this time we need that torque to provide the angular acceleration. (It's intuitively obvious that \ddot{\theta} \neq 0, but if you want to prove it, use a similar energy argument to the first part to show \ddot{\theta} = \frac{g}{2a}\sin \theta). When the wheel loses contact, there's obviously no reaction force and therefore no frictional force. So an "intermediate value argument" says at some point before losing contact, the frictional force will be insufficient to provide the necessary rotational acceleration and the wheel will slip.

Solution by DFranklin.

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