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- Jan 26, 2012

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Problem: A ball of solid sphere shape rolls without slipping down a straight inclined plane that makes an angle of $\theta$ with

the horizontal. It starts from rest at the point $x_{0}$, where $x$ is measured positively down the plane. Find the

position $x=x(t)$.

Answer. We use conservation of energy, including rotational kinetic energy, to obtain the differential equation of motion.

Let $v$ be the velocity of the ball down the inclined plane.

Let $y$ be the height of the ball at time $t$, with $h$ being the original height of the ball.

Let $m$ be the mass of the ball, and let $r$ be its radius.

Then Conservation of Energy tells us that

$$\frac{1}{2}\,mv^{2}+mgy+\frac{1}{2}\,I\omega^{2}=mgh.$$

The no-slip condition tells us that $v=\omega r$.

The moment of inertia for a solid sphere of radius $r$ is

$$I=\frac{2}{5}\,mr^{2}.$$

Hence, the energy conservation law becomes

\begin{align*}

\frac{1}{2}\,mv^{2}+mgx \sin( \theta)+ \frac{1}{2} \left( \frac{2}{5}\,mr^{2} \right) \left( \frac{v}{r} \right)^{2}&=mgh\\

\frac{1}{2}\,mv^{2}+mgx \sin( \theta)+\frac{1}{5} \,mv^{2}&=mgh\\

\frac{7}{10} \,v^{2}+gx \sin( \theta)&=gh\\

v^{2}&= \frac{10g}{7}[h-x \sin( \theta)].

\end{align*}

Let $a=\sqrt{10g/7}$, so that we obtain

$$v=a\sqrt{h-x\sin(\theta)},$$

or

$$\frac{dx}{\sqrt{h-x\sin(\theta)}}=a\,dt.$$

We can solve this separable ODE simply by integrating both sides as follows (changing $x$ to $\xi$ and $t$ to $\tau$ for

the dummy variables of integration):

$$\int_{x_{0}}^{x}\frac{d\xi}{\sqrt{h-\xi\sin(\theta)}}=a\int_{0}^{t}d\tau.$$

A simple $u$-substitution conquers the LHS integral, and the RHS integral is very straight-forward: it comes out to $at$.

Hence, we have that

$$\frac{2\sqrt{h-x_{0}\sin(\theta)}-2\sqrt{h-x\sin(\theta)}}{\sin(\theta)}=at.$$

Note that $h=x_{0}\sin(\theta)$, and thus this simplifies immediately down to

$$-\frac{2\sqrt{h-x\sin(\theta)}}{\sin(\theta)}=at.$$

Multiplying by $\sin(\theta)$ yields $-2\sqrt{h-x\sin(\theta)}=a\sin(\theta)\,t.$

Squaring both sides yields

$$4[h-x\sin(\theta)]=a^{2}\sin^{2}(\theta)\,t^{2},$$

or

$$4h-a^{2}\sin^{2}(\theta)\,t^{2}=4\sin(\theta)\,x.$$

Dividing through by $4\sin(\theta)$ yields

$$x=\frac{h}{\sin(\theta)}-\frac{a^{2}}{4}\,\sin(\theta)\,t^{2}.$$

Now $h=x_{0}\sin(\theta)$, and hence $h/\sin(\theta)=x_{0}$. Hence, we have

$$x=x_{0}-\frac{a^{2}}{4}\,\sin(\theta)\,t^{2}.$$

Plugging back in for $a$ yields

$$x=x_{0}-\frac{10g}{4\cdot 7}\,\sin(\theta)\,t^{2}=x_{0}-\frac{5g}{14}\,\sin(\theta)\,t^{2}.$$

Hence, the answer to the problem is

$$\boxed{x(t)=x_{0}-\frac{5g}{14}\,\sin(\theta)\,t^{2}}.$$

The well-known kinematic equation if you ignore rotational kinetic energy would yield

$$x(t)=x_{0}-\frac{g}{2}\,\sin(\theta)\,t^{2}.$$

So you can see that the $t^{2}$ term is off by $2/14$ or $1/7$.

Experiments using a camera to video balls rolling down inclined planes yield a typical $2\%$ error with the more accurate version, compared to over a $100\%$ error after $1$ second of rolling using the less sophisticated number.

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...g-rolling-balls-down-inclined-plane-4216.html

the horizontal. It starts from rest at the point $x_{0}$, where $x$ is measured positively down the plane. Find the

position $x=x(t)$.

Answer. We use conservation of energy, including rotational kinetic energy, to obtain the differential equation of motion.

Let $v$ be the velocity of the ball down the inclined plane.

Let $y$ be the height of the ball at time $t$, with $h$ being the original height of the ball.

Let $m$ be the mass of the ball, and let $r$ be its radius.

Then Conservation of Energy tells us that

$$\frac{1}{2}\,mv^{2}+mgy+\frac{1}{2}\,I\omega^{2}=mgh.$$

The no-slip condition tells us that $v=\omega r$.

The moment of inertia for a solid sphere of radius $r$ is

$$I=\frac{2}{5}\,mr^{2}.$$

Hence, the energy conservation law becomes

\begin{align*}

\frac{1}{2}\,mv^{2}+mgx \sin( \theta)+ \frac{1}{2} \left( \frac{2}{5}\,mr^{2} \right) \left( \frac{v}{r} \right)^{2}&=mgh\\

\frac{1}{2}\,mv^{2}+mgx \sin( \theta)+\frac{1}{5} \,mv^{2}&=mgh\\

\frac{7}{10} \,v^{2}+gx \sin( \theta)&=gh\\

v^{2}&= \frac{10g}{7}[h-x \sin( \theta)].

\end{align*}

Let $a=\sqrt{10g/7}$, so that we obtain

$$v=a\sqrt{h-x\sin(\theta)},$$

or

$$\frac{dx}{\sqrt{h-x\sin(\theta)}}=a\,dt.$$

We can solve this separable ODE simply by integrating both sides as follows (changing $x$ to $\xi$ and $t$ to $\tau$ for

the dummy variables of integration):

$$\int_{x_{0}}^{x}\frac{d\xi}{\sqrt{h-\xi\sin(\theta)}}=a\int_{0}^{t}d\tau.$$

A simple $u$-substitution conquers the LHS integral, and the RHS integral is very straight-forward: it comes out to $at$.

Hence, we have that

$$\frac{2\sqrt{h-x_{0}\sin(\theta)}-2\sqrt{h-x\sin(\theta)}}{\sin(\theta)}=at.$$

Note that $h=x_{0}\sin(\theta)$, and thus this simplifies immediately down to

$$-\frac{2\sqrt{h-x\sin(\theta)}}{\sin(\theta)}=at.$$

Multiplying by $\sin(\theta)$ yields $-2\sqrt{h-x\sin(\theta)}=a\sin(\theta)\,t.$

Squaring both sides yields

$$4[h-x\sin(\theta)]=a^{2}\sin^{2}(\theta)\,t^{2},$$

or

$$4h-a^{2}\sin^{2}(\theta)\,t^{2}=4\sin(\theta)\,x.$$

Dividing through by $4\sin(\theta)$ yields

$$x=\frac{h}{\sin(\theta)}-\frac{a^{2}}{4}\,\sin(\theta)\,t^{2}.$$

Now $h=x_{0}\sin(\theta)$, and hence $h/\sin(\theta)=x_{0}$. Hence, we have

$$x=x_{0}-\frac{a^{2}}{4}\,\sin(\theta)\,t^{2}.$$

Plugging back in for $a$ yields

$$x=x_{0}-\frac{10g}{4\cdot 7}\,\sin(\theta)\,t^{2}=x_{0}-\frac{5g}{14}\,\sin(\theta)\,t^{2}.$$

Hence, the answer to the problem is

$$\boxed{x(t)=x_{0}-\frac{5g}{14}\,\sin(\theta)\,t^{2}}.$$

The well-known kinematic equation if you ignore rotational kinetic energy would yield

$$x(t)=x_{0}-\frac{g}{2}\,\sin(\theta)\,t^{2}.$$

So you can see that the $t^{2}$ term is off by $2/14$ or $1/7$.

Experiments using a camera to video balls rolling down inclined planes yield a typical $2\%$ error with the more accurate version, compared to over a $100\%$ error after $1$ second of rolling using the less sophisticated number.

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...g-rolling-balls-down-inclined-plane-4216.html

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