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Physics 218 Final Exam Spring 2010, Sections 521-525
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Rules of the exam:
1. You have the full class period to complete the exam. 2. When calculating numerical values, be sure to keep track of units. 3. You may use this exam or come up front for scratch paper. 4. Be sure to put a box around your final answers and clearly indicate your work to your grader. 5. Clearly erase any unwanted marks. No credit will be given if we can’t figure out which answer you are
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Table to be filled by the graders
Part Score Part 1 (40) Part 2 (30) Part 3 (35) Part 4 (30) Part 5 (30) Part 6 (35) Bonus (5) Exam Total
Part 1: (30p) General questions Problem 1.1: Linear Momentum: An object is moving with constant velocity.
Question 1.1.1: (3p) Consider two inertial coordinate systems that are at rest with respect to each other. Would both coordinate systems measure the same linear momentum of the object ?
Yes, the linear momentum is the same for all coordinate systems that do not move wrt each other.
Question 1.1.2: (3p) In what situations a system of particles conserves its linear momentum?
When the sum of external forces to the system is zero.
Problem 1.2: Moment of Inertia: A weird-‐shaped solid object has mass M and moment of inertia ICM around its center of mass. Question 1.2.1: (3p) The end of a massless rod of length L is welded to the center of mass of the object. What is the moment of inertia of the rod+object system around the free end of the rod?
From the parallel axis theorem, Ifree-end-of-rod=ICM+ML2
Question 1.2.2: (3p) Objects S and H shown below have the same shape, the same mass, and the same position of their center of masses. Object S is solid and its mass is distributed homogenously throughout its volume. Object H is hollow and its mass is distributed only on the surface shell. Which object will have the larger moment of inertia with respect to the center of mass?, and why ?
Since the definition of moment of inertia involves summing of the squares of distances, the object that has the masses farther away will be have the larger moment of inertia. In this case H.
Problem 1.3: Angular Moment: an object is spinning around its center of mass. Question 1.3.1: (3p) In general, does the angular momentum of a body or system of particles depend on the inertial coordinate system in which it is evaluated?
Yes, as the definition of angular momentum depends on distances from the coordinate system.
Question 1.3.2: (3p) In general, when is the angular momentum of a system of particles conserved?
When the sum of all external torques is zero.
Question 1.3.3: (3p) Consider two inertial coordinate systems that are at rest with respect to each other. Does the angular momentum of an object spinning around its center of mass with no translational movement depend on the inertial coordinate system in which it is evaluated?
No. In this case the angular momentum can be considered intrinsic.
Problem 1.4: Torques: A motor is attached to end of an axle and provides a constant torque of τ along the axle. The other end of the axle is attached to the center of a solid disk of mass M and radius R initially at rest. Question 1.4.1: (3p) Find the magnitude of the angular acceleration of the disk around the axis. Does it depend on the time?
α = τ I
= 2τ MR2
Question 1.4.2: (3p) Is the vector of the angular acceleration parallel or transverse to the torque vector?
The vector of angular acceleration is parallel to the torque vector.
Question 1.4.3: (3p) Is the velocity vector of a particle in the rim of the disk parallel or transverse to the angular velocity?
The angular velocity is parallel to the angular acceleration which is parallel to the torque. The velocity vector of a particle in the rim is then transverse to the angular velocity.
Problem 1.5: Gravitation: Objects A and B attract each other under the force of gravity. Question 1.5.1: (3p) If the objects A and B are spherically symmetrical, what is the force that object A exerts on B and what is the force that object B exerts on A?
FA on B=-FB on A with |FBonA|=Gmamb/R2
Question 1.5.2: (3p) What is the gravitational potential energy contained in the system of A+B?
Simply U=-Gmamb/R+arbitrary constant
Question 1.5.3: (4p) If objects A and B orbit around each other following each an elliptical path. What geometrical point always lies at the focus of both ellipses?
The center of mass of the system A+B.
Part 2: (30p) Collisions Problem 2.1: A square paddle is set through axis A to a table and allowed to rotate free. The moment of inertia of the paddle around axis A is IA. Initially the paddle is spinning with an angular velocity ω 0 around axis A. A ball of mass m moving with velocity v0 towards the paddle at a distance R with respect to the axis of the paddle. The ball collisions elastically with the paddle when the paddle is exactly in the y direction as shown in the figure, and after the collision the ball moves along the x axis (positive or negative).
Question 2.1.1: (6p) Identify all the external forces acting on the ball+paddle system. Be careful, if you get this one wrong you will likely get the rest wrong.
Besides gravity and normal reactions forces, the force the axis exerts on the spinwheel at the moment of the collision.
Question 2.1.2: (5p) Is the linear momentum of the ball+paddle system the same before and after the collision?, why ?
No. The sum of external forces in the plane is not zero as the axis exerts force.
Question 2.1.3: (5p) Is any component of the angular momentum of the ball+paddle system the same before and after the collision?, why ?
Yes. Since no external force can exert torque on the z component the angular momentum on the z component must be conserved.
Question 2.1.4: (5p) Is the energy of the ball+paddle system the same before and after the collision?, why ?
Yes. The collision is elastic, and no other external forces perform work on the system.
Question 2.1.5: (9p) Write the system of equations that would in principle allow you to solve both the final velocity of the ball and the final angular velocity of the paddle. Do not solve it.
From conservation of angular momentum:
−IAω 0 +mv0R = −IAω f +mv f R
From conservation of energy:
1 2 mv0
2 + 1 2 IAω 0
2 = 1 2 mv f
2 + 1 2 IAω f
Part 3: (35p) Equilibrium Problem 3.1: A disk of radius R and mass m lies in a plane with friction and inclination α , and has a nail attached to a string as shown in the figure below. The other end of the string connects (via a pulley) to one end of a massless bridge of length L. The other end of the bridge is supported by a column and a mass 2m is located at a distance x from the column.
Question 3.1.1: (10p) Forgetting about the bridge setup, find the tension T in the string such that the disk does not move.
From the sum of forces and torques:
τz,cm = TR − fR = 0∑ ⇒ T = f Fy = N −mgcos(α)∑ = 0⇒ N = mgcos(α) Fx = mgsin(α)∑ −T − f = mgsin(α) − 2T = 0
⇒ T = mgsin(α) 2
Question 3.1.2: (8p) Forgetting about the bridge setup, find the minimum coefficient µS necessary for the disk to be in equilibrium.
We know that f ≤µsN. On the other hand from 3.1.1 we got f=T=mgsin(α)/2, so
mgsin(α)/2≤µsN=µsmgcos(α) tan(α)/2≤µs, so the coefficient must be greater than tan(α)/2
Question 3.1.3: (9p) Considering the bridge setup, find the distance x necessary to achieve equilibrium.
From sum of forces and torques over the column:
N1 +T − 2mg = 0⇒ N1 = mg 2 − sin(α) 2
−2mgx +TL = 0⇒ x = TL 2mg
= sin(α) 4
Question 3.1.4: (8p) Is that equilibrium stable? What would happen if the mass in the bridge is moved a little bit off the position of equilibrium?
If x > sin(α)*L/4 then the T would be higher than needed and the ball rolls up.
If x < sin(α)*L/4 then the T would be smaller than needed and the ball rolls down.
Therefore the equilibrium is not stable.
Part 4: (30p) Linear Kinematics Problem 4.1: A piston follows a simple harmonic motion with angular frequency ω and amplitude A. At t=0, when the piston is at its minimum, a mass m is quickly put on top of it. The piston does not changes its way of moving after the mass is added.
Question 4.1.1: (5p) Write the equation of position of the piston as a function of time for the given axis. Make sure you find all the constants from the initial conditions.
y(t)=Acos(ωt + φ). Because at t=0 the piston is at –A we get
y(t=0)=-A=Acos(φ) -1= cos(φ)φ=π
Question 4.1.2: (5p) Write the velocity and acceleration of the piston as a function of time.
v(t)=dy(t)/dt=-ωAsin(ωt + π)
a(t)=dv(t)/dt=-ω2Acos(ωt + π)
Question 4.1.3: (3p) Do the free-‐body diagram of the mass m.
Question 4.1.4: (5p) Find the normal force as a function of time.
N-mg=-m ω2Acos(ωt + π) N(t)=m(g- ω2Acos(ωt + π))
Question 4.1.5: (6p) Find the time at which the mass leaves the piston.
The mass leaves the piston when the normal is zero. Let’s call t* the time in which N(t*)=0, then
N(t*) = m g −ω 2Acos(ωt *+π )( ) = 0⇒ g ω 2A
= cos(ωt *+π )⇒ t* = Arc cos( g
ω 2A ) −π
Question 4.1.6: (6p) Find the velocity at the moment the mass leaves the piston.
Need to find the velocity at time t*,
v(t*) = −ωAsin(ωt *+π ) = −ωAsin(Arc cos( g ω 2A
Part 5: (30p) Angular Kinematics Problem 5.1: A disk is attached to a roof and free to rotate around it center axis as shown below. The disk has radius R=0.4m and non-‐homogenous mass distribution, but its center of mass coincides with the axis position. A rope is winded around the disk. The other end of the rope is attached to a mass m2=12.5 Kg (figure A) and is seen that the mass m2 accelerates downwards with acceleration am2=4.9m/s2 . A number with proper units is expected for all these questions.
Question 5.1.1: (7p) Find the moment of inertia of the disk around its axis.
Fy = T −m2g = −m2∑ am2 ⇒ T = m2(g − am2 )
τ = TR = Iα = I am2 R ⇒ I = TR
am2ext ∑ =
m2(g − am2 )R 2
am2 = 2 Kg m2
Problem 5.2: (20p) Two identical smaller objects are now glued to the side of the disk in Problem 5.1:, with their center of masses at a distance L=0.25m from the axis of the disk as shown in figure below. The two smaller objects are exactly on opposite sides of the axis, have a mass of mo=4Kg and a moment of inertia around their center of masses of Io=1 Kg m2
Question 5.2.1: (10p) Find the moment of inertia of the disk+glued objects system about the disk’s center.
The moment of inertia is the sum of the moment of inertia of the disk (calculated in 5.1.1) and the moment of inertia of each smaller object. The moment of each object is obtain just by using the parallel axis theorem:
I = 2 Kg m2 + 2 I0 +m0L 2( ) = 4.5 Kg m2
Question 5.2.2: (8p) Find the acceleration of the mass m2 in this case.
From 5.1.1 we get:
I = m2(g − am2 )R
am2 ⇒ am2 I +m2R
2( ) = m2gR2 ⇒ am2 = m2gR
I +m2R 2( )
= 3.01 m/s2
Question 5.2.3: (5p) Find the angular velocity of the disk+glued objects around its axis when the mass m2 dropped a distance h=1m.
When the mass m2 descended one meter its velocity is
v f 2 = v0
2 + 2aΔy⇒ v f = 2aΔy ≅ 2.45 m s
For that velocity the angular velocity of the disk+glued objects is:
ω f = v f R
c.m. L L
Part 6: (35p) Gravitation Problem 6.1: A small planetoid of mass m is orbiting around a much bigger sun (of mass M) on an elliptical orbit. In its orbit the distance of closest approach to the sun is Rp (perihelion), and that of farthest excursion is Ra called aphelion. Answer all the following questions in terms of m, M, Rp, Ra, and the gravitational constant G. Question 6.1.1: (8p) Using conservation of angular momentum, find the ratio of speeds at apogee and perigee.
L = mRava = mRpvp ⇒ va vP
= RP Ra
Question 6.1.2: (9p) Using conservation of energy obtain the speed of the planetoid at aphelion and perihelion.
Let’s write the equation of conservation of energy and then include the result from 6.1.1
E = 1 2 mvp
2 − GmM RP
= 1 2 mva
2 − GmM Ra
Using va = RP Ra
vP from 6.1.1 we get :
1 2 mvp
2 − GmM RP
= 1 2 m RP
Ra 2 vP
2 − GmM Ra
⇒ vp 2 −
Ra 2 vP
2 − 2GM Ra
⇒ vp 2 1− RP
⎠ ⎟ = −2GM
− 1 RP
⎠ ⎟ = vp
2 Ra 2 − RP
⎠ ⎟ = −2GM
RP − Ra RaRP
⇒ vp 2 = 2GM Ra − RP
Ra 2 − RP
2 = 2GM Ra + RP( )
⇒ vP = 2GM Ra + RP( )
The result for the aphelion velocity is then va=RpvP/Ra
Question 6.1.3: (9p) A big interstellar spaceship carrier (to be invented) wants to “tow” the planetoid far away from the sun. What is the total work the carrier performs in doing this?
The work necessary to carry the planetoid out of the sun equals the change of energy of the system. At the final destination the potential energy is zero and so is the final kinetic energy.
Wcarrier = ΔE = E f − Ei = −Ei
E = 1 2 mvp
2 − GmM RP
= 1 2 m 2GM
Ra + RP( ) Ra RP
− GmM RP
= GmM RP
Ra Ra + RP( )
⎝ ⎜ ⎜
⎠ ⎟ ⎟ =
Ra + RP( ) ⎛
⎝ ⎜ ⎜
⎠ ⎟ ⎟ = −
GmM Ra + RP( )
⇒W = −Ei = GmM Ra + RP( )
Question 6.1.4: (9p) If instead of towing the planetoid, the spaceship just slows the orbital velocity of the planetoid to zero when it is at the perihelion. What is the velocity of the planetoid when it reaches the surface of the sun ? (the radius of the sun is Rs)
Ei = Energy after the spaceship took all kinetic energy, Es = Energy at the surface of sun
Ei = Es = − GmM RP
= 1 2 mvs
2 − GmM Rs
⇒ vs 2 = 2GM 1
⎠ ⎟ ⇒ vs =
2GM RP − Rs( ) RsRP