Physics 218: Exam 3
Sections 501 to 506, 522, 524, and 526.
April 12th, 2013.
Rules of the exam:
1. You have the full class period to complete the exam.
2. Formulae are provided on the last page. You may NOT use any other formula
sheet.
3. When calculating numerical values, be sure to keep track of units.
4. You may use this exam or come up front for scratch paper.
5. Be sure to put a box around your final answers and clearly indicate your work
to your grader.
6. Clearly erase any unwanted marks. No credit will be given if we cant figure
out which answer you are choosing, or which answer you want us to consider.
7. Partial credit can be given only if your work is clearly explained and labeled.
8. All work must be shown to get credit for the answer marked. If the answer
marked does not obviously follow from the shown work, even if the answer is
correct, you will not get credit for the answer.
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I want to pick the test in the class of (circle one): 2:20 pm or 5:30pm
Question
Points
Equilibrium
24
Physics of Billiards
20
Down the hill
24
Moving Squares
28
Can of water
24
Total:
120
Score
Part 1: Equilibrium
(24 points)
A homogenous beam of mass=740 Kg is at
one end supported by a cable that hangs
from the ceiling as shown in the picture below. There is friction between the ground
and the end of the beam.
60◦
µs
40◦
(a) (8 points) Draw the free body diagram of the rod including relevant angles.
T
60◦
N
fs
40◦
mg
(b) (8 points) Find the tension in the cable. Express as a number with
proper units.
Just write the usual equations of equilibrium
X
F, x̂ :fs − T cos(60) = 0
X
F, ŷ :N − mg + T sin(60) = 0
X
L
τ, ẑ : − mg cos(40) + LT sin(60 − 40) = 0
2
from the last equation we get
mg cos(40)
T =
⇒ T = 8121 N
2 sin(20)
Page 2
(c) (8 points) Find the minimum coefficient of static friction µs between the
beam and the floor for the beam to remain in this position. Express as a
number with proper units.
From the previous set of equations we have
fs
N
from before fs − T cos(60) = 0 ⇒ fs = T cos(60) = 4061N
we need fs < µs N ⇒ µs >
from before N − mg + T sin(60) = 0 ⇒ N = mg − T sin(60)
⇒ N = 219N
therefore µs >
fs
= µs > 18.5
N
Page 3
Part 2: Physics of Billiards
(20 points)
In a billiard game the balls are solid spheres of radius R and moment of inertia
I = 25 mR2 .
(a) (10 points) Find the distance h from the ground at which the cue should
hit the ball if you want no initial friction to develop.
X
Fx :F = max
τẑ :F (h − R) = Iα, using ax = −αR we get F (h − R) = −I
ax
R
2
2
2
ax
F
= − RF
F (h − R) = − mR2 = − mR2
5
R
5
mR
5
7
2
⇒h−R= R⇒ h= R
5
5
(b) (10 points) Find the instantaneous angular acceleration of the ball if the
cue hits the ball at the center (h=R) and the coeficcient of kinetic friction
between the ball and the table is µk .
In this case the only force making a torque with respect to the center of
mass is the friction.
fk R
I
µk mgR
5µk g
α= 2
⇒ α=
2
2R
mR
5
τẑ :fk R = Iα ⇒ α =
Page 4
Part 3: Down the hill
(24 points)
40
A solid sphere of radius of R = 30cm and
mass m = 2Kg is at the top of a hill of
height h = 40m. The sphere starts rolling
down without slipping but 3/4 of the way
down it passes through an oil deposit that
removes all friction for the rest of its entire
motion.
30
20
10
00
oil
deposit
(a) (4 points) After passing the oil deposit is the angular momentum of the
ball conserved ? Explain why or why not.
Yes, because the sum of the external torques is zero.
(b) (10 points) Find the translational motion of the sphere when it reaches the
valley at the bottom of the hill.
Calling h = 40m and h1 = 10m.In going from the top to the oil deposit
we have a the force of static friction that does not work:
1
1
mgh = mgh1 + mv12 + Iw12
2
2
1 2 1 v1 2
1 12
7
mg(h − h1 ) = mv1 + I( ) = v12 (m +
m) = v12 m
2
2 R
2 25
10
10
m
⇒ v12 = g(h − h1 ) = 420( )2
7
s
Now, from that point on angular velocity does not change, so considering
conservation of energy from the point h2 to the bottom we get:
1
1
1
1
mgh1 + mv12 + Iw2 = mv 2 + Iw2
2
2
2
2
1 2 1 2
mgh1 + mv1 = mv ⇒ 2gh1 + v12 = v 2
2
2
q
√
m
v = 2gh1 + v12 = 196 + 420 = 24.8
s
(c) (10 points) Find how high up the other side of the valley will the rock go.
Simply when all the translational energy is spent on gravity:
1 2 1 2
1
mv + Iw = mghf + Iw2
2
2
2
2
v
= 31.4m
hf =
2g
Page 5
Part 4: Moving Squares
(28 points)
In a region outside any gravitational attraction a pieze of homogenous steel of
mass 3m has ”L” shape and it moves towards the right at velocity v. Another
small pieze of homogenous steel has mass m, is square, and moves towards the
first piece at velocity 2v. At the moment of the collision both pieces stick together.
2L
L
L
2v
L
2L
3m
m
L
v
(a) (4 points) Is the angular momentum of the two-piece system conserved ?.
Explain why or why not.
Yes, it is conserved b/c the sum of the external torques is zero.
(b) (4 points) Is the linear momentum of the two-piece system conserved ?.
Explain why or why not.
Yes, it is conserved b/c the sum of the external forces is zero.
(c) (4 points) Is the mechanical energy of the two-piece system conserved ?.
Explain why or why not.
No, because the work of other forces in the collision might not be zero.
(d) (4 points) Put a coordinate system and find the vertical component (ycomponent) of the position of the center of mass of the L-shaped piece.[Hint:
imagine the L-shape piece as three smaller blocks]
rcm,y
P
−m L2 + 2m L2
L
mi yi
= P
=
=
mi
3m
6
Page 6
(e) (4 points) Find the horizontal component (x-component) of the velocity of
the center of mass of the two-piece system.
vcm,x
P
3mv − m2v
mi vi
v
=
= P
=
mi
4m
4
(f) (8 points) After the pieces are stuck together find the angular velocity of
rotation of the now whole square around its center of mass.
Using conservation of angular momentum in the ẑ direction around the
center of the big square we get:
Li = Lf
L
L
1
3mv + m2v = 4mL2 ωf
6
2
6
3
4
mvL = mL2 ωf
2
6
9v
ωf =
4L
(g) (10 points (bonus)) What is the total rotational and translational kinetic
energy before and after the collision ?
1
1
7
56
Ri + Ki = 3mv 2 + m4v 2 = mv 2 = mv 2 =
2
2
2
16
1
1
2
+ Iωf2
Rf + Kf = 4mvcm
2
2
1
v 2 11
9v 2
= 4m( ) +
4mL2 (
)
2
4
26
4L
1
27
29
= mv 2 + mv 2 = mv 2
8
16
16
Page 7
Part 5: Can of water
(24 points)
faucet
A can of radius R rotates freely with initial angular velocity ω0 . The can has moment of inertia Ic around its rotating axis.
As the can is rotating somebody opens a
faucet and a dense liquid starts dropping
in the can, for a total mass of liquid of
ml . Assume the can rotates so slowly that
the liquid fills the can homogenously (i.e.
the level of water is the same for all radius
even though the can is rotating).
liquid falling
free-rotating
axis
(a) (8 points) Assuming the liquid in the can is moving like a solid disk of
radius R find the angular velocity ωf of the can filled with the liquid.
A the water drops in it exerts no torque along the direction of the freerotating axis. Since there are no torques around the free rotating axis
the angular momentum is conserved.
Li =Lf
1
Ic ω0 = (Ic + ml R2 )ωf ⇒
2
Ic ω0
ωf =
Ic + 12 ml R2
(b) (8 points) After being filled the can develops a hole in the outer edge and
the liquid starts going out. When the can is empty again, what is the
angular velocity ? If you think it did not change explain why ?
The angular velocity is still ωf , it did not change. As the liquid goes
out there is still no torque of forces on the can and therefore no angular
acceleration on it. Thus the angular velocity remains constant.
(c) (8 points) When the can is again empty due to the hole in the can, what is
the new angular momentum of the empty can?
As the liquid goes out it removes angular momentum. When the can is
again empty the angular momentum is simply L = Ic ωf .
Page 8