Sec.15.8 - Roller Derby

Applied Project in Sec.15.8, Calculus by Stewart
Chinese version 滾動競賽

Suppose that a solid ball(a marble), a hollow ball (a squash ball), a solid cylinder (a steel bar), and a hollow cylinder (a lead pipe) roll down a slope. Which of these objects reaches the bottom first?

To answer this question, we consider a ball or cylinder with mass $m$, radius $r$, and moment of inertia $I$ (about the axis of rotation). If the vertical drop is $h$, then the potential energy at the top is $mgh$. Suppose the object reaches the bottom with velocity $v$ and angular velocity $\omega$, so $v=\omega r$. The kinetic energy at the bottom consists of two parts: $\frac{1}{2}mv^2$ from translation (moving down the slope) and $\frac{1}{2}I\omega ^2$ from rotation. If we assume that energy loss from rolling friction is negligible, the conservation of energy gives $$ mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2 $$

Problem 1

Show that $$ v^2=\frac{2gh}{1+I^},\quad \text{where } I^=\frac{I}{mr^2} $$

Problem 2

If $y(t)$ is the vertical distance traveled at time $t$, then the same reasoning as used in Problem 1 shows that $v^2=2gy/(1+I^)$ at any time t. Use this result to show that $y$ satisfies the differential equation $$ \frac{dy}{dt}= \sqrt{\frac{2g}{1+I^}}(\sin{\alpha})\sqrt{y} $$ where $\alpha$ is the angle of inclination of the plane.

Problem 3

By solving the equation in Problem 2, show that the total travel time is $$ T=\sqrt{\frac{2h(1+I^)}{g\sin^2\alpha}} $$ This shows that the object with the smallest value of $I^$ wins the race.

Problem 4

Show that $I^*=1/2$ for a solid cylinder and $I^*=1$ for a hollow cylinder

Problem 5

Calculate $I^*$ for a partly hollow ball with inner radius $a$ and outer radius $r$. Express your answer in terms of $b=a/r$. What happened when as $a\to 0$ and as $a\to r$?

Problem 6

Show that $I^*=2/5$ for a solid ball and $I^*=2/3$ for a hollow ball. Thus the objects finish in following order: solid ball, solid cylinder, hollow ball, hollow cylinder.

Answer

Solid ball is the situation $a \to 0$ in Problem 5, thus $I^*=2/5$

Hollow ball is the situation $a \to r$ in Problem 5, thus $I^*=2/3$

According to Problem 3, the object having smaller $I^*$ will arrive sooner.

Since $2/5 < 1/2 < 2/3 < 1$, the objects arrive in following order: solid ball, solid cylinder, hollow ball, hollow cylinder.