Problem #1: PART A: Calculate the center-of-mass of the array of point masses indicated below. Use the XY coordinate system indicated on the diagram. The smaller two masses are m=2 kg each and the two largest masses are M=3 kg. The four masses are connect by very light, rigid pieces of wire. PART B: Calculate the moment of inertia about the vertical axis.
PART C: Calculate the moment of inertia about the horizontal axis.
2 Exam Review Problems answers copy.nb
ANSWER PART A:
2 * H-0.5L + 2 * H1.0L + 3 * H-0.5L + 3 * H1.0L X = 2 + 2 + 3 + 3 0.25
2 * H0.25L + 2 * H0.25L + 3 * H-0.25L + 3 * H-0.25L Y = 2 + 2 + 3 + 3 -0.05
ANSWER PART B:
Iv = 2 * H-0.5L2 + 2 * H1.0L2 + 3 * H-0.5L2 + 3 * H1.0L2 6.25
ANSWER PART C:
Ih = 2 * H0.25L2 + 2 * H0.25L2 + 3 * H-0.25L2 + 3 * H-0.25L2 0.625 Exam Review Problems answers copy.nb 3
Problem #2: A cylinder having a moment of inertia I=(1/2)MR2 rolls down an incline as pictured below:
Calculate the angular velocity of the cylinder when it reaches the bottom of the inclined plane. Take H = 7 m, M = 0.25 kg, R = 0.5 m, and q = 30°. Do NOT use Newton's laws of motion. 4 Exam Review Problems answers copy.nb
Conservation of energy yields KE(translation)+ KE (rotation) = PE e(it top of inclined plane)
1 1 2 2 MV + Icylinder w = M g H (1) 2 2 Using the connection between angular velocity w and translational velocity V=wR or w = V/R is used together with the 2 moment of ineretial of a cylinder Icylinder=(1/2)MR yields
1 1 MV2 + IH1 ê 2L M R2M HV ê RL2 = M g H (2) 2 2
1 1 3 4 MV2 + MV2 = M g H and thus V2 = g H and finally V = gH (3) 2 4 4 3
Substitution of the numbers for g=9.8 m/s2 and H=7m yields for the linear velocity V= 9.6 m/s .
4 V = * 9.8 * 7 3
9.56382 and for the angular velocity w=4.8 Rad/sec.
w = 9.6 ê 0.5
19.2
Problem #3: A figure skater can increase her spin rotation rate from an initial rate of 0.75 Rev every 2 seconds to a final rate of 2 3.5 Rev/sec. If her initial moment of inertia was I0=4.0 kg-m what is her final moment of inertia I f ? She physically accom- plishes this increase rotation rate by moving her arms closer to her body.
Conservation of angular momentum says that L f = L0 where L0 is the intial angular momentum of the skater and L f is the final angular momentum of the skater when she has her hands closer to her body. The initial angular momentum of the skater is 2 L0 = I0 w0 when the skater's arms are outstretched with an initial moment of intertia I0 = 4.0 kg - m and the skater is rotating with an angular velocity w0 . Note that w0=2p f0 with the initial rotation rate given by f0=0.75Rev/2sec=0.375 Rev/sec. The final angular momentum of the skater is L f = I f w f when the skater's arms are are closer to her body with an initial moment of 2 intertia I f = 4.0 kg - m and the skater is rotating with an angular velocity w f . Note that w f =2p f f with the initial rotation rate given by f f =3.5 Rev/2sec. So L f = L0 means that I0 w0 = I f w f and solving for w f we get the final momentum I f = I0 2 ( w0/ wf)=0.43 kg-m .
4.0 * H2 * p * 0.375L ê H2 * p * 3.5L
0.428571 Exam Review Problems answers copy.nb 5
Problem #4: Consider a painter on a ladder as indicated in the picture below. The mass of the ladder is m=7 kg and the mass of the painter is M=62 kg. Suppose the coefficient of friction is 0.6 between the floor and ladder. Assume there is no friction force between the ladder and the wall. How high up the ladder can the painter climb in the y-direction? Can the painter climb up the maximum possible of Y=4.0 m?
Y
X
First notice that the length of the ladder L= 32 + 42 = 5 m and that Cos[q]=3/5=0.6
1. Equilibrium requires the sum of the forces in the x direction to be zero
‚ Fx = FCx - FW = 0 (4)
0.6 * 676.
405.6
This says that FCx = FW so if we know one, then we know the other. 2. Equilibrium also requires the sum of the forces in the y direction to be zero
‚ Fy = FCy - Mg - mg = 0 (5)
2 This allows us to calculate FCy = Mg + mg = H62 kg + 7 kgL 9.8m/s = 676.2 Nt
3. The law of friction say FCx = mFCy=0.6*676.2=405 Nt and thus FW =405 Nt.
4. The sum of the torques must also be zero for the ladder to be in equilibrium
‚ t = FW H4 mL - FCyH0 mL - FCxH0 mL - Mg X - mg 2.5 Cos@qD = 0 (6)
Use bottom end of the ladder as the reference point for calculating torque. The shortest distance from the line of force mg of the ladder is (L/2)*Cos[q]=(5/2)*0.6=2.1 Also call X the shortest distance form the line of force Mg to the reference point. The shortest distance is 0 meters from the line of force FCy to the reference point at the bottom end of the ladder. Similarly, the shortest distance is 0 meters from the line of force FCx to the reference point at the bottom end of the ladder. Simplifcation of equation (6) yields
405*4 = 62*9.8* X + 7*9.8*2.1 (7) and solving for X yields 6 Exam Review Problems answers copy.nb
X = H405 * 4 - 7 * 9.8 * 2.1L ê H62 * 9.8L
2.42913
Since X is less than the distance 3 m between the bottom of the ladder and the wall, you can conclude the painter cannot climb to the top of the ladder without it slipping. The distance D along the ladder the painter can climb is D=4.04 m given by
D = X ê Cos@qD
D = X ê 0.6
4.04855
By the way, the angle q is about 53° as can be obtained from the calculation
q = [email protected] * 180 ê p
53.1301
Note the ladder is 5 meters long so this is less than the length of the ladder.
The height Y the painter can climb on the ladder is given by
Y = X * Tan@53 °D
3.22357
So the maximum height in the y direction the painter can climb is about 3.22 meters out of a possible 4 meters.
Problem #5: A 1200 kg boxcar traveling at 15.0 m/s strikes a second boxcar at rest. The two stick together and move off with a speed of 6.0 m/s. What is the mass of the second car? How much kinetic energy is lost in the collision?
15 m/sec
at rest
Before Collision
6 m/s
After Collision
Using conservation of linear momentum
1200 kg * 15 m ê s = H1200 kg + ML * 6.0 m ê s (8) and solving for M we get Exam Review Problems answers copy.nb 7
M = H1200 * 15 - 1200 * 6L ê 6
1800
So the other railroad car has mass 1800 kg. The initial kinetic energy is 135,000 J
1 KE0 = * 1200 * 152 2. 135 000. while the final kinetic energy is 54,000 J
1 KEf = * H1200 + 1800L * 62 2. 54 000. so 81,000 J of energy is lost in the collision.
KE0 - KEf
81 000.
Problem #6: Two balls of equal mass masses mA = 0.04 kg and mB = 0.04 kg are suspended as show in the diagram below.
The ball mA is pulled away to a 60° angle with the vertical and released. (a) What is the velocity of the ball mA just before impact? (b) What will be the maximum height ball mB after the elastic collision? Describe what happens to ball mA as a result of the collision.
ü From when ball mA is released until just before mA hits mB.
Energy is conserved for this process so m gh= 1 m V2 and thus V = 2 g h where h=L-L Cos[60°] and this works out to A 2 A h=0.15 m (see below) and V=1.71 m/s. (This is the answer to PART A)
h = 0.30 - 0.30 * Cos@60 °D
0.15 8 Exam Review Problems answers copy.nb
g = 9.8; V = 2 * g * h
1.71464
ü Assume momentum and energy is conserved during the collision process:
f f f f mA V = mAVA + mBVB thus V = VA + VB since mA=mB. 2 2 2 2 1 m V2 = 1 m IV f M + 1 m IV f M thus V2 = IV f M + IV f M These two equations imply V f V f = 0. 2 A 2 A A 2 B B A B A B f f CASE #1: The solution VB = 0 means that VA = V which does NOT make sence physically since mA would pass through mB. f f CASE #2: The solution VA = 0 means that VB = V This case makes physical sense. In this case, mass A stops and mass B rises up a height h on the right and side of the appratus. (This is the answer to PART B.) The final height of ball B is 0.15 m.
Problem #7: Calculate the angular velocity of the Earth as it orbits the Sun and give your answer in Rad/sec. The known fact is that the period for the Earth about the Sun is T=365 days/Rev.
T = 365; f = 1 ê T; w = 2 * p * f ê H24. * 60 * 60L
1.99238 µ 10-7
ANSWER: The angular velocitiy w is 2.0 ä 10-7 Rad/sec
Problem #8: A wheel 0.5 meter in diameter accelerates uniformly from 120 rpm to 370 rpm in 3.0 seconds. How far will a point on the edge of the wheel have traveled in meters in this time?
The initial frequency 120 Rev/Min*(Min/60 Sec) so the initial angular velocity is w0=2*p*(120./60) and the final frequency is wf=2*p*(370./60) Since this happens in t=3.0 seconds you can compute the angular acceleration a=8.73 Rad/sec2 . The angular distance traveled is q=77 Rad which works out to a linear distance S=Rq=38.4 meters.
w0 = 2 * p * H120. ê 60L; wf = 2 * p * H370. ê 60L; t = 3.0 ; a = Hwf - w0L ê t 8.72665
The angular distance traveled q in that time is given by
q = w0 * t + H1 ê 2L * a * t2 76.969
The corresponding linear distance S traveled is
R = 0.25; S = R * q 19.2423
Problem #9: A rotating merry - go - round makes one complete revolution in 6.0 seconds (a) What is the linear speed of a child seated 1.4 meter from the center? (b) What is her radial acceleration? 2 2 ANSWER: v=1.47 m/s and aR = V ëR = 1.54 mësec
T = 6.0; f = 1 ê T; w = 2 * p * f
1.0472 Exam Review Problems answers copy.nb 9
R = 1.4; V = w * R 1.46608
aRadial = V2 ë R 1.53527
You can also calculate the radial acceleration using the formula below and of course you get the same answer
aRadial = w2 * R 1.53527
Problem #10: How much work is required to stop an proton that has a mass m=1.67 ä 10-27 kg and is moving with a speed 1.90 ä 106 of m/s? ANSWER: Work = 2.83ä10-15 Joules (see below using Work= DKinetic Energy Theorem)
mp = 1.67 * 10-27; V = 1.9 * 106; KEf = 0; KE0 = H1 ê 2L * mp * V2; Work = KEf - KE0
-3.01435 µ 10-15
Problem #11: Newton's second law of motion is not valid in a rotating coordinate system. The centrifugal force is a fictional force which is included in the usual Newton's second law of motion so it works in rotating coordinate systems. The centrifugal force is in the outward radial direction while the centripetal force is in the inward radial direction. The centripetal force is a real force like for example gravity.
TRUE or FALSE (Circle One)
Problem #12: Con Edison sells electricity by charging a $0.119 per KiloWattHour. The KiloWattHour a unit of
POWER ENERGY (Circle the correct answer)
Comment: The usual unit of energy is the Joule and the usual unit of power is Watt=Joule/sec. If you multiply P ä t =W where t is the time the power is used, then you get the energy W. So a Watt-Sec=Joule.
Problem #13: Radians and Degrees are two different units used to measure angular distance q. Both the Radian and the Degree measurement of an angle are the same regardless of the size of the radius of the circular motion.
TRUE or FALSE (Circle One) 10 Exam Review Problems answers copy.nb
I meant to say that "the degree measure of an angle is independent of the size of the radius of the circular motion AND so too is the radian measure of an angle independent of the size of the radius of the circular motion."
Problem #14: Suppose a mass slides down a hill as pictured below. Suppose the M=5 kg mass is released from rest at the top of the hill which is a height H=3.5 meters above the reference level. How fast is the mass going when it reach the bottom of the hill which is h=1.7 meters above the reference level? Assume there is no friction force. Do not use Newton's laws of motion to solve this problem.
Inital Position M
Final Position H
h Reference Level
ANSWER: M g H = 1 M V2 +M g h so V= 2 g HH - hL =5.93 m/sec 2
g = 9.8; H = 3.5 ; h = 1.7; V = 2 * g * HH - hL
5.9397
Problem #15: Consider the point P on the circular object which is undergoing rotational motion and is pictured below. The point P is a fixed distance R from the center of the circle. Suppose the tangential velocity of the point P increases. Circle all of the following that are TRUE. (a) The angular velocity increases. (b) The angular acceleration is not zero. (c) A torque must have been applied to the circle. (d) A tangential force must have been applied to the circle. (e) The radial acceleration increases.
ANSWER: All the parts of this question are TRUE and should be circled. Exam Review Problems answers copy.nb 11