Question

A solid disk and a hoop are simultaneously released from rest at the top of an incline and roll down without slipping. Which object reaches the bottom first? The one that has the largest radius arrives first.

Solution

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We know that solid disk and a hoop are simultaneously released from rest at the top of an incline and roll down without slipping.

The question is which will object will reach the bottom first.

Let's start with energy conservation equation:

12mv2+12Iω2mgh=0\dfrac{1}{2}\cdot m\cdot v^2+\dfrac{1}{2}\cdot I\cdot \omega^2-m\cdot g\cdot h=0

Moment of inertia of solid disc is given by:

I=12mr2I=\dfrac{1}{2}\cdot m\cdot r^2

Moment of inertia of hoop is given by:

I=mr2I=m\cdot r^2

Angular velocity is given by:

ω=vr\omega=\dfrac{v}{r}

We can use energy equation so find term for velocity of solid disc:

12mv2+1212mr2(vr)2mgh=0\begin{align*} \dfrac{1}{2}\cdot m\cdot v^2+\dfrac{1}{2}\cdot \dfrac{1}{2}\cdot m\cdot r^2\cdot \left(\dfrac{v}{r}\right)^2-m\cdot g\cdot h&=0\\ \end{align*}

Solving it for velocity:

v=4gh3v=\sqrt{\dfrac{4gh}{3}}

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