Math 1321
Week 12 Lab Worksheet
Due Thursday 04/04
1. Applied Project - Roller Derby: 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. Each of these objects will have the same radius R and mass m. Which of
these objects reaches the bottom first? (Make a guess before proceeding.)
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 ω, so v = ωR. The kinetic energy at the bottom consists of two
parts: 12 mv 2 from translation (moving down the slope) and 12 Iω 2 from rotation. If we
assume that energy loss from rolling friction is negligible, then conservation of energy
gives
1
1
mgh = mv 2 + Iω 2
2
2
(a) Show that
v2 =
2gh
1 + I∗
where
I∗ =
I
mR2
Solution:
1
1
mgh = mv 2 + Iω 2
2
2
1
= (m + I/R2 )v 2
2
2mgh
v2 =
m + I/R2
2gh
=
1 + I∗
(b) if y(t) is the vertical distance traveled at time t, then the same reasoning as used
in (a) shows that v 2 = 2gy/(1 + I ∗ ) at any time t. Use this result to show that y
satisfies the differential equation
dy
=
dt
r
2g
√
(sin α) y
∗
1+I
where α is the angle of inclination of the plane. Hint: The vertical component of
= v sin(α)
speed is dy
dt
Solution: The vertical component of the speed is v sin α so,
r
dy
2gy
sin α
=
dt
1 + I∗
r
2g
√
=
sin α y
1 + I∗
(c) By solving the differential equation in (b), show that the total travel time is
s
T =
2h(1 + I ∗ )
g sin2 α
Solution: Solving the separable differential equation, we get
dy
√ =
y
r
2g
sin α dt
1 + I∗
√
2 y=
r
2g
(sin α)t + C
1 + I∗
But y = 0 when t = 0, so C = 0. Solve for t when y = h
√
2 y=
r
2g
(sin α)t
1 + I∗
√ s
2 h 1 + I∗
T =
sin α
2g
s
2h(1 + I ∗ )
=
g sin2 α
RRR 2
(d) Show that I ∗ = 1/2 for a solid cylinder. (Hint: Iz =
(x + y 2 )ρ(x, y, z) dV ,
E
ρ = m/(πR2 l), where ρ is density, m is mass and l is the length of the cylinder).
Solution: Assume that the length of each cylinder is l. Then the density of
the solid cylinder is m/(πR2 l).
ZZZ
m
(x2 + y 2 )ρ(x, y, z) dV
2
πR l
Z l Z 2π Z R
m 2
r r dr dθ dz
=
2
0
0 πR l
0
h 1 iR
m
=
2πl r4
2
πR l
4 0
mR2
=
2
Iz =
and so, I ∗ = Iz /(mR2 ) = 1/2.
(e) Show that I ∗ = 1 for a hollow cylinder. Hint: Use the same formula for Iz from
part(d) but for a different density function ρ(x, y, z), where all of the mass of the
cylinder is located on the outer cylindrical shell or perimeter.
Solution: For the hollow cylinder, we consider its entire mass to lie a distance
R from the axis of rotation, so ρ(x, y, z) = 0 unless r2 = x2 + y 2 = R2 in which
m
. Thus the computation of the moment of inertia simplifies
case ρ(x, y, z) = 2πRl
to a double integral of the form
ZZZ
m
ρ(x, y, z)dV
πR2 l
Z l Z 2π Z R
ρ(r, θ, z)r2 rdrdθdz
=
0
0
0
Z Z
mR3 l 2
=
πdθdz
2πRl 0 0
= mR2
Iz =
so, I ∗ = Iz /(mR2 ) = 1
(f) It can be shown that I ∗ = 2/5 for a solid ball, and I ∗ = 2/3 for a hollow ball.
Finally, since we have shown I ∗ = 1/2 for the solid cylinder, I ∗ = 1 for the hollow
cylinder, which object wins the race? Which one places second, third, and fourth?
Solution: From part (c), we saw the total length of time it takes for an object
with moment of inertia I ∗ to travel the length of the ramp is given by
s
2h(1 + I ∗ )
T =
g sin2 α
Thus, the object with the moment of inertia which gives the smallest value of T
will win the race. By this argument, the solid ball will win the race as it has the
smallest moment of inertia. The remaining finish in the following order: solid
cylinder, hollow ball, hollow cylinder.