Download PHY 116 From Newton to Einstein Model Answers to Exercise Sheet 5

PHY 116 From Newton to Einstein
Model Answers to Exercise Sheet 5: Rigid Bodies
Marks out of 30
1).
A turntable has a kinetic energy of 0.0250 J when turning at 45.0 rev/min.
What is the moment of inertia of the turntable about the axis of rotation?
[1]
[1]
[1]
[1]
2
E
0
.
05
2
=
Answer: E = ½ Iω 2 ⇒ I = 2 =
0
.
00225
kg
m
ω
(45 × 2π / 60)2
2).
Small blocks, each with a mass m, are clamped at the ends and at the centre of
a light rod of length L. Calculate the moment of Inertia of the system about an axis
perpendicular to the rod and passing through a point ¼ of the length from one end.
You can ignore the moment of inertia of the rod.
[1]
[3]
[1]
2
2
2 2
2
3
L
L
3 L
11L
Answer: I = ∑ mi ri2 = m 2 + m 2 + m 2 =
16
4
4
4
i =1
3).
A flywheel with a radius of 0.3m starts from rest and accelerates with a
constant angular acceleration of 0.6 rad/s2. Compute the magnitude of the tangential
acceleration, the radial acceleration, and the resultant acceleration of a point on the
rim a) at the start; b) after it has turned through 60° and c) after it has turned through
120°.
[1]
[2]
2
2
Answer: aT = αr , a R = rω = r (αt ) = 2rαθ
(a)
At t = 0, θ = 0,
aT = 0.6 × 0.3 = 0.18 m s −2
aR = 0
(b)
[1]
[1]
a = aT = 0.18 m s −2
At θ = 60°,
aT = 0.6 × 0.3 = 0.18 m s −2
a R = 2 × 0.3 × 0.6 × π / 3 = 0.377 m s −2
(c)
[1]
a = aT2 + a R2 = 0.418 m s −2
At θ = 120°,
aT = 0.6 × 0.3 = 0.18 m s −2
a R = 2 × 0.3 × 0.6 × 2π / 3 = 0.754 m s −2
a = aT2 + a R2 = 0.775 m s −2
[1]
4).
A cord is wrapped around the rim of a wheel 0.25m in radius, and a steady
pull of 40.0 N is exerted on the cord. The wheel is mounted on a frictionless bearing
on a horizontal shaft through its centre. The moment of inertia of the wheel about the
shaft is 5.00 kg.m2. Compute the angular acceleration of the wheel.
Answer:
[2]
Torque τ = Fr = 10 N m.
τ 10
Also τ = Iα, so α = =
= 2 rad s −2 [3]
5
I
5).
Under some circumstances a star can collapse into an extremely dense object
made mostly of neutrons and called a neutron star. The density of the neutron start is
roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the
start as a uniform solid rigid sphere, both before and after collapse. The star’s initial
radius was 7.0 x 105 km (comparable to the Sun); its final radius is 16 km. If the
original star rotated once in 30 days, find the angular speed of the neutron star.
Answer:
Angular momentum L is conserved, and L = Iω = c MR2ω where c is
some factor appropriate to the sphere. So
[1]
cMR12ω1 = cMR22ω 2
[2]
2 [1]
5 2
 7 × 10 
R
2π

ω 2 = ω1 12 = 
= 4639 rad s -1
R2  16  30 × 24 × 60 × 60
6).
Occasionally, a rotating neutron star undergoes a sudden speedup called a
glitch. One explanation is that a glitch occurs when the crust of the neutron start
settles slightly, decreasing the moment of inertia about the rotation axis. A neutron
star with angular speed ω0 = 70.4 rad/s underwent such a glitch in October 1975 that
increased its angular speed to ω = ω0 + Δω, where Δω/ω0 = 2.01 x 10-6. If the radius
of the neutron star before the glitch was 11 km, by how much did the radius decrease
in the starquake? Assume that the neutron star is a uniform sphere.
Answer:
Using
cMR12 ω1 = cMR22 ω 2
R22 = R12
ω1
ω2
[1]
[1]
[1]
 ∆ω0 
ω0
∆ω0
ω1

= R1
= R1 1 −
= R1 1 −
R2 = R1
ω2
ω0 + ∆ω0
ω0
2ω0 

∆ω0 [1]
∆R = R1 − R2 = R1
= 11 km × ½ × 2.01 × 10 −6 = 11 mm [1]
2ω 0