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A small, light ball and a larger, heavier ball are released from the top of a slope.

Which will move further? which will come down faster?

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Like this it looks too much like a homework. What does confuse you in this problem? Why you can not solve it yourself? – Maksim Zholudev Jan 15 '12 at 18:52

2 Answers 2

Assumptions: The slope is not slippery, the balls have similar mass distribution, negligible air drag.

The angle of the slope is $\theta$, and mass of one of the ball is $m$, and its radius is $R$.

$a=\alpha R$ where $\alpha$ is the angular acceleration of the ball, and torque by static friction $f$ is $\tau = f R$, and $ mg \sin \theta - f = ma$. we also know $\tau = \beta mR^2 \alpha$ where $\beta$ is some constant (depends only on the distribution of mass).

Solving for $a$ yields $a=\frac{g \sin\theta}{1+\beta}$, independent of $m$ and $R$. Therefore, both balls should have same acceleration (and therefore same velocity and displacement).

However, if there's air drag, their acceleration depends on the radius and the mass of the ball, so not enough information is given for this case. If the balls are of the same material (same density $\rho$), the larger one comes down faster than the smaller one. Here's a proof...

After very long time, since air drag is proportional to the velocity $f_d=6\pi\eta v R$ ($\eta$ = viscosity of air), the balls will eventually reach their terminal velocities (denoted as $v_t$) and their accelerations are close enough to zero. With setting $f_d = 6\pi\eta v_t R = m g \sin\theta = \frac43 \pi \rho R^3$, we get the fact that the terminal velocity is proportional to the square of the radius. Therefore the larger one comes down faster.

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Air drag depends on the surface area and the velocity$^2$. The larger ball will experience slower acceleration and will be behind when the small ball reaches terminal velocity. It would need to be a very long slope for the larger ball to catch up and pass the smaller ball. – LDC3 Mar 22 at 16:31

If friction and air resistance are negligible(if you want them in the calculations,just say it because now i am just assuming that this is the simplest problem) and the spheres start rolling from the same point,then the law f conservation if energy states that Mgh=0.5MV^2.You have M in both sides,so you can see that the final velocity is not depended on the mass of the object.So they will have the same velocity. This can also be explained with Newton's laws.Putting calculations aside and just using intuition,we know that two bodies of different mass have the same acceleration,so having them slide in an angle does not change anything(it certainly changes things,but it changes nothing as far as the acceleration and the intuition go)!

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