Worksheet 9: Rigid-Body Rotation
Answers
1. The equatorial radius of the earth is about 6400 km. It completes one rotation per stellar
day, which is 23 hours, 56 minutes, and 4.1 seconds.
a. What is the angular speed ω in rad/s about the polar axis of a point on Earth’s surface
at a latitude of 40° N?
Period T = 34h + 56 min + 4.1 s = 82800 s + 3360 s + 4.1 s = 86164.1 s
Angular speed ω = 2π/T = 7.29211×10−5 rad/s
b. What is the linear speed v of that point? Assume that the earth is a perfect sphere (not
true, but pretty close).
θ
y r
v = ωy where y = r cos θ
v = ωr cos(40°) = (7.29211×10−5 rad/s)(6.4×106 m)(0.7660) = 357.51 m/s
c. What is the centripetal acceleration of that point?
a = ω2r = ωv = (7.292×10−5/s)(357.51 m/s) = 0.026 m/s2
2. A steam turbine runs with a constant angular velocity of 150 rpm. When steam is shut
off, the friction of the bearings and the air stops the rotor in 2.2 h.
a. What is the constant angular acceleration, in rev/min2, of the rotor during its slowdown?
Δt = 2.2 h·(60 min/h) = 132 min; Δω = −150 rev/min
α = Δω/Δt = (−150 rev/min)/(132 min) = −1.136 rev/min2
b. How many rotations does the rotor make before stopping?
Δθ = ωt + ½ αt2 = (150 rev/min)(132 min) + ½ (−1.136 rev/min2)(132 min)
=19800 rev − 9896.8 rev = 9903.2 rev
(Should be exactly half as many as if it continued at its initial speed.)
c. At the moment that the turbine is turning at 75 rpm, what is the tangential component
of the linear acceleration of a flywheel particle that is 50 cm from the axis of rotation?
a = αr = (−1.136 rev/min2)(0.50 m/rad)(2π rad/rev) = 3.57 m/min2·(min/60 s)2
= 9.91×10−4 m/s2
x=0
3. What is the moment of inertia of an isosceles triangle of
mass M, height A, and base B, rotated about its axis of
symmetry?
y=0
a. First find the area density σ = mass/area of the
triangle. Obtain a formula in terms of the quantities
M, A, and B.
Area = ½ AB so σ =
+x
+y
A
dy
2M
AB
B
b. Using the coordinates and origin shown in the diagram to
the right, find an expression for x, the distance of a point on
the edge of the triangle from the axis, in terms of y.
x=
B
y
2A
c. We’ll evaluate the moment of inertia by adding together (integrating) the moments of
inertia of strips between ±x with width dy rotated about their centers. What is the
mass dm of such a strip? The formula should be in terms of σ, x or y, and dy.
dm = σ(2x)dy = 2σxdy = σ
B
ydy
A
d. What is the moment of inertia dI of such a strip? Recall that I of a thin rod rotated
about its center is ML2/12.
dI =
1
12
2
dm(2x) =
2
1
12
B
⎛ B ⎞
σ ydy ⎜ y ⎟ =
A
⎝ A ⎠
3
1
12
⎛ B ⎞
σ ⎜ y ⎟ dy
⎝ A ⎠
e. Now find the total moment of inertia of the triangle by integrating ∫dI. You’ll want to
make the variable of integration y, evaluated from y = 0 to y = A.
I = ∫ dI =
1
12
B3 A 3
B3 y 4
1
σ 3 ∫ y dy = 12 σ 3
A 0
A 4
y= A
=
y =0
1
48
B3 4
σ 3A =
A
1
48
σAB 3 .
f. Substitute in the formula for σ, so that I is in terms of no more than M, A, and B.
I=
1
48
2M
AB 3 =
AB
1
24
MB 2 . Notice no dependency on A.
g. Compare the result to the moment of inertia of a rectangle of axial length A,
transverse width B, and mass M. Does it make sense?
The rectangle has I = ML2/12. This is half that, which makes sense, because
more of the mass is closer to the axis in the triangle.