Download PHYSICS 218

PHYSICS 218 Honors
EXAM 2 Retest
Choose 5 of the following 6 problems. Indicate which problem is not to be graded.
1. A rope is affixed at one end to the rim of a pulley, and wrapped five turns around the
pulley. The pulley has a mass of 5 kg, a radius of 0.2 m, and its mass is distributed as a
uniform disk. A mass of 10 kg is suspended from the loose end of the rope, so that it
hangs 2 m from the floor. The mass is then released. What is the velocity of the mass
just before it strikes the floor?
Solution: The rope will unwind by an angle
2m
θ = L/ R =
= 10rad
0.2m
We require Newton's law for force and for torque: (M=10 kg is mass of weight, m=5 kg
is mass of pulley)
Mg − T = Ma
TR = Iα
a = Rα
mR 2
2
mRα
Mg −
= Ma
2
Mg
a=
= 8m / s 2
M + m/2
I=
Now solve the equation of motion:
1
x = at 2
2
t = 2x / a
v = at = 2ax = 5.6m / s
2. A lawn roller in the form of a thin-walled hollow cylinder of mass M is pulled
horizontally with a constant force F applied by a handle attached to its axle. If it rolls
without slipping, find the acceleration and the friction force.
Solution:
There will be a friction force Ff opposing the external force F, so as to increase the
angular velocity as the linear velocity increases (no slipping). Require Newton’s law for
both forces and torques:
F − F f = Ma
F f R = Iα
a = Rα
I = Mr 2 (thin cylindrical shell, all mass on the surface (radius R)
F − Iα / R = Ma
F − Ma = Ma
a = F / 2M
F f = Ma = F / 2
3. Consider a uniform solid disk of radius 25 cm, with a mass of 20 kg. Three 10 cm
diameter holes are bored through the disk as shown. What is the moment of inertia of the
holey disk, about its axis?
Solution: Use the parallel axis theorem:
The moment of inertia of the big disk with no holes, about its axis, is
1
I disk = MR 2 = 0.625 kg m 2
2
The moment of inertia of each small disk (radius r, distance from center of big disk d,
mass m = Mr2/R2) that is removed, about the central axis, is
.008 .009
1 2


2
I hole = mr + md = .001 + .018 = .019
2
.032 .033


3
So overall, I = I disk − ∑ I holen = 0.565 kg m2
n =1
4. Two astronauts must retrieve a satellite that has begun to tumble in orbit. The satellite
is in the shape of a long cylinder, whose mass of 1,000 kg is uniformly distributed inside.
The satellite is tumbling end over end, with a period of 5 minutes. The astronauts want to
stop the tumbling by catching the satellite at its ends. If they each apply constant force
over a maximum stroke of 0.3 m, how much force must they apply to stop the satellite?
Solution:
Call the overall length of the cylindrical satellite L.
The moment of inertia of a rod rotating about its center in a tumbling motion is
1
I = ML2
12
L
The torque exerted by the two astronauts is τ = 2( ) F
2
The angular acceleration is then
α = τ / I = 12 F / ML
2π
The initial angular velocity is ω =
= .021rad / s
5 min
The angle through which the ship rotates before stopping is
1
θ = ω 0 t − αt 2
2
ω = ω 0 − αt
ω = 0 when θ =
ω 02
= .018 L / F
2α
We require that the angle θ correspond to a stroke s = 0.3 m.
L
s = ( )θ = .009 L2 / F = 0.3
2
F = .03L2
5. Communications satellites are placed in geosynchronous orbit: the satellite orbit
remains over a fixed location on the Earth’s surface. Calculate the radius of a
geosynchronous orbit. What is the highest latitude on Earth’s surface that can receive
line-of-sight signals from such satellites?
The universal constant of gravitation is 6.67 x 10-11 N m2/kg2.
The mass of the earth is 6 x 1024 kg.
Hint: you don’t need these, only g and the radius of the Earth, 6400 km, and the length of
a day!
Solution:
For geosynchronous orbit, we must have that the orbit period equals one day:
2π
ω=
= 7.3 ⋅ 10 −5 rad / s
24(3600s )
Apply Newton’s law to require that the acceleration from gravity equals that needed to
keep it in orbit:
R2
F = GMm / r 2 = GM / RE2 m 2E
14243 r
g
(
)
v2
F =m
= mω 2 r
r
r =
3
gRE2
ω
2
(9.8m / s )⋅ (6.4 ⋅ 10 m)
=
(7.3 ⋅ 10 / s )
2
6
−5
2
2
= 7.5 ⋅ 10 24 m 3
r = 1.9 ⋅ 10 8 m
The highest latitude that can receive line-of-sight signals is an angle β from the pole,
R
6.4 ⋅ 10 6 m
= .034rad = 2°
where sin β = E =
r
1.9 ⋅ 10 8 m
So the highest latitude is 90o-2o = 88o.
6. A uniform pulley wheel 8 cm diameter, mass 1 kg, has a 5 m long cord wrapped
around its periphery. Starting from rest, the wheel is given an angular acceleration of
1.47 rad/s2. A) through what angle must the wheel turn for the cord to unwind? B) how
long does it take? C) what is the final angular momentum of the wheel?
Solution:
A) θ = L / R =
5m
= 62.5rad
.08m
1
B)
2
t = 2θ / α = 9.2 s
1

L = Iω =  MR 2 ω
2

C) ω = αt = 111rad / s
θ = αt 2
L = .08kgm 2 / s