A student watches two people coast down a hill on bikes. They both start from rest. Neither one of them peddles the bike. Gravity is the ONLY driving force. Friction and air resistance are negligible. Of the two people, the person with the larger mass reaches the bottom of the hill significantly earlier than the person with less mass. Explain why this happens.
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3$\begingroup$ It's not possible. In the absence of friction both experience the same downhill acceleration $a=g\sin\theta$, with $\theta$ the angle of the slope with respect to the horizontal. $\endgroup$– GertCommented Feb 20, 2016 at 22:33
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$\begingroup$ Perhaps the person with more mass started further down the hill. This is a brain teaser, not physics. $\endgroup$– Jon CusterCommented Feb 20, 2016 at 23:29
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$\begingroup$ An approach to thinking about this is to ask where the potential energy goes: some of it goes into translational kinetic energy, but some of it goes somewhere else: where, in the absence of frictional losses? $\endgroup$– user107153Commented Feb 21, 2016 at 0:39
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$\begingroup$ @Gert - this is not true, see en.wikipedia.org/wiki/… $\endgroup$– YukterezCommented Feb 21, 2016 at 3:17
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1$\begingroup$ @СимонТыран: sure, if you take $I$ of the wheels into account but let's try and stay serious, please. ;-) $\endgroup$– GertCommented Feb 21, 2016 at 4:03
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2 Answers
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Assuming they have identical bicycles with wheels having moment of inertia $I$ and radius $R$, their velocity after dropping a height $H$ is given by
$$ v^2=\frac{2\cdot g\cdot H}{1+\frac{I}{m\cdot R^2}}$$
Thus the larger $m$ has a larger velocity.
answered Feb 21, 2016 at 2:17
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$\begingroup$ please note that friction cannot be negligible, otherwise the bicycles would skid, not roll. Friction depends on the normal force and the complete problem should take it into account, $\endgroup$– anna vCommented Feb 23, 2016 at 6:48
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This could be explained if the wheels on the bicycle ridden by the person with the larger mass had a smaller moment of inertia.
If the wheels have a moment of inertia then the loss of gravitational potential energy is equal to the gain in translational kinetic energy of the centre of mass bicycle and rider plus the gain in rotational kinetic energy of the wheels.
For a larger moment of inertia the proportion of rotational kinetic energy increases and so the proportion of translational kinetic energy ($\propto v^2_{\textrm{translational}}$) decreases.
