Download Exam 3

Name & ID
University of Alabama
Department of Physics and Astronomy
PH 125 / LeClair
Spring 2014
Exam III Solutions
1. A wad of sticky clay with mass m and velocity vi is fired at a solid cylinder of mass M and radius R as shown
below. The cylinder is initially at rest and mounted on a fixed horizontal axle that runs through its center of mass.
The line of motion of the projectile is perpendicular to the axis and at a distance d < R from the center. Find the
angular speed of the system just after the clay strikes and sticks to the surface of the cylinder. The moment of
inertia of a solid cylinder is I = 21 M R2 , the moment of inertia of a point particle mass m a distance R from an
axis of rotation is I = mR2 .
m vi
M
R
d
Solution: Clearly this is an inelastic collision, since the wad of clay sticks to the cylinder, and conservation of
energy is right out. Conservation of momentum is fine, but the cylinder rotates after the collision and makes
things difficult. Conservation of angular momentum will be a whole lot easier. Since we must eventually find the
angular velocity of the cylinder, it is natural to consider angular momentum about the center of the cylinder.i
Consider first the moment just before the wad hits the cylinder. The wad has momentum p = mv, acting at a
distance R from the center of the cylinder. The angular momentum about the center of the cylinder is the wad’s
linear momentum times the perpendicular distance to the axis of rotation, which is just d. The angular momentum
about the center of the cylinder is thus Li = mvd.
After the wad hits the cylinder and both begin to rotate with angular velocity ω, the angular momentum can be
found from the moments of inertia and ω. We have the wad, with Iw = mR2 , and the cylinder, with Ic = 21 M R2 ,
both rotating with angular velocity ω, and thus the total angular momentum after the collision is
1
1
Lf = Iw ω + Ic ω = mR ω + M R2 ω = R2 ω m + M
2
2
2
Equating initial and final angular momentum, we can easily find ω:
1
R2 ω m + M
2
= mvd
ω=
i
mvd
R2
m + 21 M
It does not really matter, angular momentum will be conserved regardless of our choice of origin. Choosing the center of the
cylinder is simply more convenient.
Name & ID
2. A uniform disk with mass M = 2.5 kg and radius R = 20 cm is mounted on a fixed horizontal axle, as shown
below. A block of mass m = 1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find
the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. Note: the
moment of inertia of a disk about its center of mass is I = 12 M R2 .
M
m
Solution: In order to get acceleration and angular acceleration, we’ll need to use force and torque, respectively.
Start with the pulley. The tension T in the rope pulls on the edge of the disk a distance R from the center of
rotation at an angle of θRT = 90◦ , which causes a torque τ . This torque must equal the disk’s moment of inertia
times the angular acceleration.
1
τ = RT sin θRT = RT = Iα = M R2 α
2
2T
α=
MR
(1)
(2)
We can get the tension by considering the force balance for the hanging mass. We have the tension in the tope
pulling up, the weight of the mass pulling down, and an overall acceleration downward. Thus
X
F = T − mg = −ma
(3)
Noting that a = Rα, this gives T = mg−M Rα. Now we’ve got two equations for α, which we can combine.
2T
2
2mg 2m
=
(mg − mRα) =
−
α
MR
MR
MR
M
2mg
2m
=α 1+
MR
M
2mg
≈ −24 rad/s2
α=
R (M + 2m)
α=
(4)
(5)
(6)
Given α, we can find a and T .
2mg
≈ −4.8 m/s2
M + 2m
2m2 g
2m
mM
T = mg − M Rα = mg −
= mg 1 −
=g
≈ 6.0 N
M + 2m
M + 2m
M + 2m
a = Rα =
(7)
(8)
Name & ID
3. A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 0.60 m, and
the density of iron is 7870 kg/m3 . If the density of water is 1000 kg/m3 , find the inner diameter of the sphere.
Solution: The shell will just barely float when its weight is equal to the buoyant force. The buoyant force is
equal to the weight of the displaced water, and is thus represented by a volume of water equal to the volume of
the whole shell, whose outer radius we will call ro (let the inner radius be ri ). This volume times the density of
water times g gives the buoyant force.
FB = ρw
4 3
πr g
3 o
(9)
The weight of the shell is the density of the iron times its volume times g. The volume of iron in the shell would
be the volume of a sphere of radius ro minus the volume of a sphere of radius ro .
FW = ρi
4 3 4 3
πr − πr g
3 o 3 i
(10)
Balancing the two forces,
0 = FB − FW = ρw
4 3 4 3
4 3
πro g − ρi
πr − πr g
3
3 o 3 i
ri3 ρi = ro3 (ρi − ρw )
s
ri = ro
3
(11)
(12)
ρi − ρw
= ro
ρi
r
3
1−
ρw
≈ 0.287 m
ρi
(13)
The inner diameter is thus about 0.573 m, giving the shell a wall thickness of about 3 mm.
4. A (spherical) star of radius R = 5×105 km has a rotational period of 60 days. Later in its life, its radius expands
to R = 5 × 106 km, though its mass M remains constant. What is the new rotational period after expansion?
Presume the star’s moment of inertial is kM R2 at all times.
Solution: We can’t readily use conservation of energy here, because to do so we’d have to account for the energy
it took to accomplish the expansion. One thing we can rely on, however, is conservation of angular momentum.
Both before and after, the angular momentum L = Iω should be the same. We only need to know the relationship
between angular velocity ω and period T . In one period T , the star rotates through 2π radians at angular velocity
ω, so it must be the case that
ω=
∆θ
2π
=
∆t
T
(14)
Presuming the star’s mass remains constant and only its radius changes, conservation of angular momentum gives
Ii ωi = If ωf
(15)
kM Ri2 ωi = kM Rf2 ωf
Ri2
2π
Ti
= Rf2
Tf =
Rf
Ri
2π
Tf
(16)
!
(17)
2
Ti = 6000 days
(18)
Name & ID
5. A small body of mass m hangs in equilibrium at one end of a light string of length l, the upper end of which
√
is fixed. A small body of mass m moving horizontally with velocity 2 gl strikes the former body and adheres to
it. Find:
(a) the velocity with which the combined bodies begin to move,
(b) the angle through which the string turns before coming to rest for an instant,
Solution: Conservation of momentum will get us the velocity after the collision.
p
mvi = 2m gl = 2mvf
=⇒
vf =
p
gl
(19)
If the string turns through an angle θ, its endpoint is now a distance l cos θ from the vertical plane from which it
hangs, rather than a distance l away as it is before the collision. The mass has then risen by an amount l−l cos θ.
The maximum angle will be determined by the balance of the potential energy gained by moving through this
height and the original kinetic energy after the collision.
1
(2m) vf2 = 2mg∆y
2
mgl = 2mgl (1 − cos θ)
2 − 2 cos θ = 1
1
cos θ =
2
θ = 60◦
(20)
(21)
(22)
(23)
(24)
6. Ethanol of density ρ = 791 kg/m3 flows smoothly through a horizontal pipe that tapers in cross-sectional area
from A1 = 1.20 × 10−3 m2 to A2 = A1 /2. The pressure difference between the wide and narrow sections of pipe is
4120 Pa. What is the volume flow rate (i.e., m3 /s) RV of the ethanol?
Solution: Bernoulli’s equation is the one thing we know that can relate pressure and fluid speed. It is essentially
energy conservation, and states that pressure plus kinetic energy plus gravitational potential energy should have
the same value everywhere for an ideal fluid. Applied to the present case, this sum should be the same at both
ends of the pipe.
1
1
p1 + ρv12 + ρgy1 = p2 + ρv22 + ρgy2
2
2
(25)
Here ρ is the same on both sides if we assume the ethanol is essentially incompressible. Since the pipe is horizontal,
y1 = y2 , and we can simplify.
1 p1 − p2 = ρ v22 − v12
2
(26)
We still have the velocity to deal with. We are after the volumetric flow rate Rv = Av, which involves velocity and
the known areas, so we can substitute for the velocities in favor of known quantities and the flow rate we want to
Name & ID
find.
Rv
A1
Rv
2Rv
v2 =
=
A2
A1
!
2
1
Rv
4Rv2
p1 − p2 = ρ
− 2
2
A21
A1
v1 =
s
Rv = A1
2 (p2 − p1 )
≈ 2.24 × 10−3 m3 /s
3ρ
(27)
(28)
(29)
(30)