Download Final Spring 2011 with solutions

Physics 218 Final Exam
Fall 2011, Sections 513,514,515,526,528 Do not fill out the information below until instructed to do so! Name :____SOLUTIONS____ Signature :______________________________ Student ID :______________________________ E-­‐mail :______________________________ Section # :______________________________ Rules of the exam: 1.
2.
3.
4.
5.
You have the full class period to complete the exam. When calculating numerical values, be sure to keep track of units. You may use this exam or come up front for scratch paper. Be sure to put a box around your final answers and clearly indicate your work to your grader. Clearly erase any unwanted marks. No credit will be given if we can’t figure out which answer you are choosing, or which answer you want us to consider. 6. Partial credit can be given only if your work is clearly explained and labeled. 7. 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. Put your initials here after reading the above instructions:__________ 1 Part 1: (35p) Collisions Table to be filled by the graders Problem 1.1: A flat object of irregular shape, mass m, and moment of inertia Icm around an axis perpendicular to its plane and passing through its center of mass is hanging at rest from a nail located a distance d above its center of mass. A projectile with negligible mass but linear momentum P hits the objects horizontally at a distance L below the hanging point where it gets absorbed immediately into the object. In the following the system refers to the object+projectile system. Question 1.1.1: (4p) Identify the nail forces external to the system. Part Score Part 1 (35) Part 2 (35) Part 3 (30) Part 4 (35) Part 5 (30) Part 6 (35) Bonus (10) Exam Total Forces are gravity and whatever force the nail is exerting. L P d c.m. Question 1.1.2: (4p) During the collision is the linear momentum of the system conserved? Explain why. No, because the sum of the external forces is not zero, in particular the nail may exert a horizontal force during the collision. Question 1.1.3: (4p) During the collision is the total angular momentum conserved with respect to the point at the nail? Explain why. Yes, because the sum of the external torques in the z direction is zero and therefore the angular momentum is conserved in the z direction. Question 1.1.4: (4p) During the collision is the mechanical Energy of the system conserved? Explain why. No, because the work of non-­‐conservative forces such as those acting during the collision remove mechanical energy from the system. Question 1.1.5: (4p) Right after the collision is the mechanical Energy of the system conserved? Explain why. Yes, because there are not any non-­‐conservative forces left after the collision. Question 1.1.6: (7p) Find the angular velocity of the object right after the collision with the projectile. LP
Li = LP = (I CM + md 2 )! f ! ! f =
(I CM + md 2 )
of mass of the object will raise. Question 1.1.7: (8p) Find how much the center 1
1
1
L2 P 2
Ei = (I CM + md 2 )! 2f = mg!y " !y =
(I CM + md 2 )! 2f =
(I CM + md 2 )
2
2mg
2mg
(I CM + md 2 )2
1
L2 P 2
!y =
2mg (I CM + md 2 )
2 Part 2: (35p) Angular Momentum Problem 2.1: I the middle of space a massless support structure holds two axes. Around those axes two disks with radii R1 and R2 and masses m1 and m2 respectively are rotating without slipping against each other as shown in the figure below. The angular velocity of disk 1 is w1. While the disks are rotating around themselves, the two disks are not rotating around each other (i.e. the massless structure is not rotating around itself.) Question 2.1.1: (5p) Find the distance from the axis of R1 the bigger disk (R1) to the center of mass of the system. R2 Question 2.1.2: m 0 + m2 (R1 + R2 )
rcm = 1
=
m1 + m2
Question 2.1.3: (10p) Find the moment of inertia of the whole system around its center of mass, ISYSTEM. I SYSTEM = I1 + m1rcm2 + I 2 + m2 (R1 + R2 ! rcm )2 Question 2.1.4: (20p) After some time friction forces on the axes slows the disks rotation until they do not rotate around their own axes anymore. Find the angular velocity of rotation of the system around its center of mass. Conservation of angular momentum implies I1!1 ! I 2! 2 = I SYSTEM ! S
and because both disk rotate without slipping we get R1!1 = R2! 2 " ! 2 =
R1!1
R2
#
R &
!1 % I1 ! I 2 1 (
#
R2 '
R!
R &
$
" I1!1 ! I 2 1 1 = !1 % I1 ! I 2 1 ( = I SYSTEM ! S " ! S =
R2
R2 '
I SYSTEM
$
3 Part 3: (35 points) Energy and Collisions Problem 3.1: Two identical objects of mass m are released from rest in a smooth hemispherical bowl of radius R, from the positions shown in the figure below. You can ignore friction between the masses and the surface of the bowl. Question 3.1.1: (20p) If they stick together when they collide, how high above the bottom of the bowl will the masses go after colliding? At the moment the left mass reaches the one at
the bottom it has a speed of
1
Ei = mgR = mvB2 ! vB = 2gR
2
Because the linear momentum is conserved during
the collision and just after the velocity of the system is
vB
gR
=
2
2
Again, using conservation of energy we found
v = 2gR = mvB =2mvA ! vA =
1
v2
1 gR R
2mvA2 = 2mg"y ! "y = A =
=
2
2g 2g 2
4
Question 3.1.2: (9p) Find the ratio of the total mechanical energy at their maximum position to the mechanical energy before the collision. R
E f 2mg 4 1
=
=
Ei
mgR
2
Question 3.1.3: (6p) Is the collision elastic or inelastic? Explain your reasoning. Since the Mechanical energy is not conserved the collision is inelastic. 4 Part 4: (35p) Equilibrium Problem 4.1: A pole is used to support a sign of uniform mass m and length L. One side of the sign is hinged to the pole and the other side connected to a rope making an angle α which is also attached to the pole as shown in the figure below. Question 4.1.1: (17p) Find the Tension on the rope g From the equation of rotational equilibrium:
L
mg
T sin(! )L ! mg = 0 " T =
2
2sin(! )
α
m L Question 4.1.2: (18p) Find the horizontal and vertical component of the force on the hinge. From the other equations of equilibrium we get: mg
mg
cos(! ) =
2sin(! )
2 tan(! )
mg mg
N y + T sin(! ) ! mg = 0 " N y = mg ! T sin(! ) = mg !
=
2
2
N x ! T cos(! ) = 0 " N x = T cos(! ) =
Question 4.1.3: (0p) Knowing that the pole will bend if at any point on the pole the sum of the external torques exceeds τ max. Find the maximum height of the hinge above ground such that the pole does not break. By a mistake in the phrasing you should ignore this question. Just for fun seen as a whole, the mass of the sign will produce a torque on the pole of mgL/2, regardless of the total height of the pole. Notice that also the torque of the tension in the upper part of the pole with respect to the hinge is Tcos(alpha) *Ltan(alpha) =Tsin(alpha) = mgL/2. So as long as the τ max is larger than mgL/2 the pole will ot break. 5 Part 5: (30p) Gravitation Problem 5.1: You are designing a satellite to move in a circular orbit around the earth. You know the mass of the earth me=5.97 1024 Kg and the gravitational constant G=6.67 10-­‐11 m3 Kg-­‐1 s-­‐2. In the following you can ignore the gravitational pull of the sun and any other planets. Question 5.1.1: (10p) Write a free-­‐body diagram of the satellite and using Newton’s laws deduce the period of rotation of the satellite as a function of distance from the center of the earth. !
Gme ms
v2
4! 2 R
1
Gme
4! 2 R 3
=
!m
=
m
"
=
"
T
=
s
s
R2
R
T2
T 2 4! 2 R 3
Gme
Question 5.1.2: (10p) The Department of Defense asks you find the orbital radius necessary such that the satellite takes 24 hours to orbit the earth, that way the satellite will always be atop the same position on the surface of the earth (known as a geosynchronous orbit). What radius is that? A number with proper units is required. T=
4! 2 R 3
Gme
! R = 3 T2
" 42, 000km Gme
4! 2
Question 5.1.3: (10p) What would be the speed of the satellite? A number with proper units is required !
Gme ms
v2
Gme
Gme
=
!m
"
= v2 " v =
# 3,071 m/s s
2
R
R
R
R
6 Part 6: (35p) Simple Harmonic Motion of a planetoid Problem 6.1: A small object of mass m is attached to two springs with constants k and negligible natural lengths that in turn attach to two different concrete blocks. The mass is originally a distance h with respect to an imaginary line joining the two blocks as shown in the figure below. Assume the distance L is much larger than h. No gravity is present. Question 6.1.1: (10p) Draw a coordinate system and find the total force that both strings exert on the mass. Indicate the direction of the force. Choosing positive +X in the horizontal right position. We get
L h m In the direction of each spring we have a force of F = !k( h 2 + L2 ).
The sum of forces in the direction of x we have
Fx = 2sin(! )F = !2
L h
2
2
h +L
k h 2 + L2 = !2hk
Problem 6.2: (10p) Using the equations of Newton write the differential equation whose solution is the equation of motion. By using f=ma we get Fx = !2hk = m
d 2h
2k d 2 h
"
!h
=
dt 2
m dt 2
Problem 6.3: (10p) Find the angular frequency of oscillation of the mass between the two blocks, and write the position of the mass as a function of time. !=
2k
m
2k
x(t) = h cos(t
)
m which is are equivalent to the equations of a spring with k’=2k.
Problem 6.4: (5p) Find the velocity of the mass as it passes through the imaginary line. Seeing that the energy of spring is just Knowing that the energy of a spring in SHO is
1
1 2
2kh 2
E = k 'h 2 = kh 2 = mvmax
! vmax =
2
2
m
7