03 midterm RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm
Gravity
^ F21 = -G m12m2 r12 ,
r12
1
for r R,
g(r) = G M r2
G = 6.67259 10-11 N m2 /kg2 Rearth = 6370 km, Mearth = 5.98 1024 kg
2 Circular orbit: ac = v = 2 r = r
2 2 r = g(r) T M U = -G mr , E = U + K = - G mrM 2 2 F = - d U = -m G M = -m v 2 r dr r r0 r2 = 1- = 2L = const. m
1 Intensity: I = P = 2 v ( smax )2 A I Intensity level: = 10 log10 I , I0 = 10-12 W/m2 0 Plane waves: (x, t) = c sin(k x - t) c Circular waves: (r, t) = sin(k r - t) 1 v Doppler effect: = v T , f0 = T , f = Here v = vsound vobserver , is wave speed relative to moving observer and = (vsound vsource )/f0 , detected wave length established by moving source of frequency f0 . freceived = fref lected c Spherical: (r, t) = r sin(k r - t) r
s P = -B V = -B x V Pmax = B smax = v smax x Piston: m = m At = A v t V
Kepler's Laws of planetary motion: r0 0 i) elliptical orbit, r = 1- rcos r1 = 1+ ,
1 ii) L = r m r - A = 2 r r t t t
21 r1 +r2 , T 2 = 4 2 r 3 iii) G M = 2 T a a, a = 2 GM a2 Escape kinetic energy: E = K + U (R) = 0
Fluid mechanics
Pascal: P = F1 = F2 , 1 atm = 1.013 105 N/m2 A1 A2 Archimedes: B = M g, Pascal=N/m2 P = Patm + g h, with P = F and = m A V
1 Shock waves: Mach Number= vsource = sin vsound
Superposition of waves
h F = P dA - g 0 (h - y) dy Continuity equation: A v = constant Bernoulli: P + 1 v 2 + g y = const, 2
P 0
Phase difference: sin(k x - t) + sin(k x - t - ) Standing waves: sin(k x - t) + sin(k x + t) Beats: sin(kx - 1 t) + sin(k x - 2 t) Fundamental modes: Sketch wave patterns String: = , Rod clamped middle: = , 2 2 Open-open pipe: = , 2
Oscillation motion
1 f = T , = 2 T 2 2 x S H M: a = d 2 = - 2 x, = d 2 = - 2 dt dt x = xmax cos( t + ), xmax = A v = -vmax sin( t + ), vmax = A a = -amax cos( t + ) = - 2 x, amax = 2 A E = K + U = Kmax = 1 m ( A)2 = Umax = 1 k A2 2 2 Spring: m a = -k x Simple pendulum: m a = m = -m g sin Physical pendulum: = I = -m g d sin Torsion pendulum: = I = -
Open-closed pipe: = 4
Temperature and heat
Conversions: F = 9 C + 32 , K = C + 273.15 5 Constant volume gas thermometer: T = a P + b 1 Thermal expansion: = 1 d , = V d V dT dT = T , A = 2 A T , V = 3 V T Ideal gas law: P V = nRT = N kT R = 8.314510 J/mol/K = 0.0821 L atm/mol/K k = 1.38 10-23 J/K, NA = 6.02 1023 , 1 cal=4.19 J Calorimetry: Q = c m T, Q = L m First law: U = Q - W , W = P dV Conduction: H = Q = -k A T , Ti = -H ki t A i W Stefan's law: P = A e T 4 , = 5.67 10-8 m2 K 4
Wave motion
Traveling waves: y = f (x - v t), y = f (x + v t) In the positive x direction: y = A sin(k x - t - ) 1 T = f , = 2 , k = 2 , v = = T T k
F fixed end: phase inversion Reflection of wave: open end: same phase General: E = K + U = Kmax P = E = 1 m (A)2 2 t t Waves: m = m x = m v t x t x 1 P = 2 v ( A)2 , with = m x Circular: m = m A r = m 2 r v t A r dt A Spherical: m = m 4 r2 v t V
Kinetic theory of gas
Along a string: v =
v F = px = md x t N 2 Pressure: P = N F = m N vx = mV v 2 A V 3 1 P = 2 N K, K x = K = 2 k T , T = 273 + Tc , 3 V 3 P V = N k T , n = N/NA , k = 1.3810-23 J/K, NA = 6.02214199 1023 #/kg/mole Constant V: Q = U = n CV T Constant P: Q = n CP T
Ideal gas: px = 2 m vx ,
2
Sound
v=
B,
s = smax cos(k x - t - )
= C P , CP - C V = R V CV = d R, for transl.+rot+vib, d = 3 + 2 + 2 2 Adiabatic expansion: P V = constant 1 Mean free path: = (v vrms t t d2 n = 1 2 )
rel rms V
C
2 d nV
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm
2
Mechanics - Basic Physical Concepts
Math: Circle: 2 r, r 2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = -b 2 a-4 a c
Cartesian and polar coordinates: y x = r cos , y = r sin , r 2 = x2 + y 2 , tan = x Trigonometry: cos cos + sin sin = cos( - ) sin + sin = 2 sin + cos - 2 2 cos + cos = 2 cos + cos - 2 2 sin 2 = 2 sin cos , cos 2 = cos2 - sin2 1 - cos = 2 sin2 2 , 1 + cos = 2 cos2 2 Vector algebra: A = (Ax , Ay ) = Ax^ + Ay i ^ Resultant: R = A + B = (Ax + Bx , Ay + By ) Dot: A B = A B cos = Ax Bx + Ay By + Az Bz Cross product: ^ = k, k = ^, k ^ = i ^ ^ ^ ^ i ^ i ^ ^ i Ax Bx ^ Ay By ^ k Az Bz
B F s = F s cos = F s A F ds
(in Joules)
Effects due to work done: Fext = m a - Fc - fnc Wext |AB = KB - KA + UB - UA + Wdiss |AB B Kinetic energy: KB -KA = A m ads, K = 1 m v 2 2
B K (conservative F ): UB - UA = - A F ds 1 k x2 Uspring = 2 Ugravity = m g y, From U to F : Fx = - U , Fy = - U , Fz = - U x y z
Fspring = - U = -k x x U = 0, 2 U > 0 stable, < 0 unstable Equilibrium: x x2 W Power: P = ddt = F v = F v cos = F v (Watts)
Fgravity = - U = -m g, y
Collision
C =AB =
d d 1 Calculus: dx xn = n xn-1 , dx ln x = x , d sin = cos , d cos = - sin , d d d dx const = 0
C = A B sin = A B = A B , use right hand rule
Measurements
Impulse: I = p = pf - pi tif F dt Momentum: p = m v x Two-body: xcm = m1m1 +m2 x2 1 +m2 pcm M vcm = p1 + p2 = m1 v1 + m2 v2 Fcm F1 + F2 = m1 a1 + m2 a2 = M acm K1 + K2 = K1 + K2 + Kcm Two-body collision: pi = pf = (m1 + m2 ) vcm vi = vi - vcm , vi = vi + vcm Elastic: v1 - v2 = -(v1 - v2 ), vi = -vi , vi = 2 vcm - vi Many body center of mass: rcm = Force on cm: Fext = dp = M acm , dt
mi r i = mi r dm mi
t
Dimensional analysis: e.g., 2 F = m a [M ][L][T ]-2 , or F = m v [M ][L][T ]-2 r N (a x + b) = a N x + b N Summation: i i=1 i=1 i
p=
pi
Rotation of Rigid-Body
Motion
One dimensional motion: v = d s , a = d v dt dt s -s v -v Average values: v = tf -tii , a = tf -tii f f One dimensional motion (constant acceleration): v(t) : v = v0 + a t +v s(t) : s = v t = v0 t + 1 a t2 , v = v02 2 2 = v2 + 2 a s v(s) : v 0 Nonuniform acceleration: x = x0 + v0 t + 1 a t2 + 2 1 j t3 + 1 s t4 + 1 k t5 + 1 p t6 + . . ., (jerk, snap,. . .) 6 24 120 720 Projectile motion: trise = tf all =
ttrip v0y 2 = g 1 g t2 , R = v t h = 2 f all ox trip 2 1 Circular: ac = v , v = 2 T r , f = T (Hertz=s-1 ) r 2 + a2 Curvilinear motion: a = at r
Isphere = 2 M R2 , Ispherical shell = 2 M R2 5 3 I = M (Radius of gyration)2 , I = Icm + M D2 Kinetic energies: Krot = 1 I 2 , K = Krot + Kcm 2 Angular momentum: L = r m v = r m r = I Torque: = d L = m d v r = F r = I d = I dt dt dt Wext = K +U +Wf , K = Krot + 1 m v 2 , 2 P =
s Kinematics: = r , = v , = at r r 2 Moment of inertia: I = mi ri = r2 dm 2 2 Idisk = 1 M R2 , Iring = 1 M (R1 + R2 ) 2 2 1 M 2, I 1 M (a2 + b2 ) Irod = 12 rectangle = 12
Rolling, angular momentum and torque
I Rolling: K = 1 Ic + M R2 2 = 1 Rc + M v 2 2 2 2 Angular momentum: L = r p, L = r p = I
Relative velocity: v = v + u
Law of Motion and applications
Force: F = m a, Fg = m g, F12 = -F21
2 Circular motion: ac = v , v = 2 T r = 2 r f r Friction: Fstatic s N Fkinetic = k N
Torque: = d L = r d p = r F , = r F = I dt dt 1 Gyroscope: p = d = L d L = L = m g h I dt dt
Static equilibrium
Equilibrium (concurrent forces):
i Fi = 0
Energy
Work (for all F): W = WAB = WB - WA
i = 0 Fi = 0, about any point Subdivisions: rcm = mA rAcm +mB rBcm mA +mB Elastic modulus = stress/strain stress: F/A strain: L/L, x/h, -V /V
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm Version number encoded for clicker entry: V1:1, V2:1, V3:3, V4:3, V5:2. Question 1 part 1 of 1 10 points Water flows through a 0.246 m radius horizontal pipe at the rate of 0.188 m3 /s. The pressure in the pipe is atmospheric. This pipe is connected to a second horizontal pipe with a radius of 0.181 m, positioned 0.847 m lower than the first pipe. The acceleration of gravity is 9.81 m/s2 . What is the gauge pressure in the lower pipe? 1. 5929.56 Pa 2. 6114.19 Pa 3. 6303.37 Pa 4. 6503.9 Pa 5. 6706.07 Pa 6. 6914.13 Pa 7. 7129.71 Pa correct 8. 7357.24 Pa 9. 7585.22 Pa 10. 7821.2 Pa Explanation: Basic Concepts: P1 + 1 2 1 2 v1 + g h 1 = P 2 + v2 + g h 2 2 2 A = r 2 flow rate = A v Given: r1 = 0.246 m flow rate = 0.188 m3 /s P1 = P 0 r2 = 0.181 m h2 - h1 = 0.847 m = 1.00 103 kg/m3 g = 9.81 m/s2 Solution: v1 = flow rate flow rate = 2 A1 r1 v2 = flow rate flow rate = 2 A2 r2 1 2 2 (v1 - v2 ) + g(h1 - h2 ) 2 1 2 flow rate
2
3
P2 - P 1 = =
1 1 4 - r4 r1 2
+ g(h1 - h2 ) = (1000 kg/m3 ) 1 2 0.188 m3 /s 1 1 - 4 (0.246 m) (0.181 m)4
2
+ (9.81 m/s2 )(0.847 m) = 7129.71 Pa Question 2 part 1 of 1 10 points A student of mass 51 kg wants to walk beyond the edge of a cliff on a heavy beam of mass 320 kg and length 5.2 m. The beam is not attached to the cliff in any way, it simply lies on the horizontal surface of the clifftop, with one end sticking out beyond the cliff's edge:
d
The student wants to position the beam so it sticks out as far as possible beyond the edge,
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm but he also wants to make sure he can walk to the beam's end without falling down. How far from the edge of the ledge can the beam extend? 1. 2.175 m 2. 2.24259 m correct 3. 2.31864 m 4. 2.3964 m 5. 2.4727 m 6. 2.55353 m 7. 2.64 m 8. 2.7251 m 9. 2.81324 m 10. 2.90196 m Explanation: Basic principle: An unsecured body supported from below remains stable if and only if its center of gravity projects down inside the area supporting the body. If the body in turn supports something else above it, then the overall center of gravity (of the body plus everything on top of it) must project down inside the supporting area. For the problem in question, the stability of the beam -- with the student standing on it -- requires that the center of gravity of the beam + student system must lie above the cliff itself. If this overall center of gravity were to move beyond the cliff's edge, the beam would tilt down and fall off the cliff, taking the student with it. Let the cliff's edge be the origin of our coordinate system. The beam extends from X1 = d - L < 0 (on the cliff) to X2 = d > 0 (off the cliff), so assuming the beam is uniform, its center of mass is located at
beam Xc.m. =
4
student system is therefore located at m overall student Xc.m. = Xc.m. m+M M beam Xc.m. + m+M L M . = d - m+M 2 The stability condition is that this overall center of mass should stay on the cliff, so
overall Xc.m. < 0
and hence d < dmax = L M = 2.24259 m . m+M 2
Question 3 part 1 of 1 10 points A child takes a ball and places it at the top of a slide, at an initial height. The ball starts from rest, rolls without slipping down the slide and flies off the end into the air, with an initial angle of 32 . Ignore loss to friction and air resistance. Which statements are true throughout the ball's motion? I. Angular momentum of the ball must be conserved. II. Linear momentum of the ball must be conserved. III. Mechanical energy of the ball must be conserved. IV. The ball will keep traveling upward until it reaches its initial height. V. The ball will travel up to a maximum height less than its initial height. 1. II and IV only 2. III and IV only 3. III and V only correct 4. I, II and IV only 5. I and IV only 6. II and V only 7. I, II, III and IV only
L X1 + X 2 = d - . 2 2
When the student walks all the way to the beam's end, his own center of mass is located at
student Xc.m. d,
where the approximation is neglects the size of the student compared to the beam's length. The overall center of mass of the beam +
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm 8. I, II, III and V only 9. II, III and V only 10. I and V only Explanation: Because no energy is lost or dissipated, the total mechanical energy of the system is conserved. Neither the linear, nor angular momentum are conserved because there are external forces and torques on the ball, respectively. The ball does not reach it's initial height because a portion of its energy is locked in rotational kinetic energy and horizontal kinetic energy. Question 4 part 1 of 1 10 points A uniform disk of radius 0.2 m is mounted on a frictionless, horizontal axis. A light cord wrapped around the disk supports a 1.5 kg object, as shown in the figure. When released from rest the object falls with a downward acceleration of 2.2 m/s2 . The acceleration of gravity is 9.8 m/s2 . 9. 0.248508 kg m2 10. 0.256398 kg m2 Explanation: Let : M = mass of disk r = 0.2 m , m = 1.5 kg , a = 2.2 m/s2 , and g = 9.8 m/s2 . Basic Concepts: = d at = dt r
5
=rF =I Solution: Define down to be positive; i.e., the direction of the acceleration of the mass. Newton's 2nd law gives mg - T = ma T = mg - ma. The net torque = I is given by Tr=I a . r
Hence, substituting in the expression for the tension, T , and solving for I gives 0.2 m M T 2.2 m/s2 I = m [g - a] r2 a = (1.5 kg) [(9.8 m/s2 ) - (2.2 m/s2 )] (0.2 m)2 (2.2 m/s2 ) mg -ma = I a r2
1.5 kg What is the moment of inertia of the disk? 1. 0.188027 kg m2 2. 0.194688 kg m2 3. 0.20102 kg m2 4. 0.207273 kg m2 correct 5. 0.215436 kg m2 6. 0.223521 kg m2 7. 0.230912 kg m2 8. 0.24 kg m2
= 0.207273 kg m2 . Question 5 part 1 of 1 10 points A 3.61 m long rod of negligible weight is attached one on end to a ball joint which allows the rod to rotate in all directions. The free end of the rod points 26 above the Eastward
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm horizontal direction. A 42.5 N force directed vertically Down is applied at the rod's free end. What is the direction of the torque due to this force (relative to the pivot end of the rod)? 1. Up and Southwest 2. Horizontally Westward 3. Horizontally Northward correct 4. The torque vanishes 5. Vertically Up 6. Insufficient data to determine the direction. 7. Horizontally Southward 8. Vertically Down 9. Horizontally Eastward 10. Down and Eastward Explanation: Use the right hand rule for = X F : Stick your right thumb parallel to the rod -- diagonally upward of East -- and let your fingers point in the direction of the force -- vertically Down. Consequently, the plain of your hand is vertical and your palm faces horizontally Northward. Alternatively, you may use a right coordinate system: the x axis points East, the y axis point North and the z axis points Up. ^ In terms of the unit vectors ^, ^ and k, we i ^ acting at the location have a force F = -F k ^ X = (L cos )^ - (L sin ) k. Consequently, i the torque ^ ^ = -F L cos (^ k) + F L sin (k k) i ^ = -F L cos (-^) + F L sin (0) = +(F L cos ) ^ points along the positive y axis horizontally Northward. =XF 6.9 kg 50 cm 4.9 kg 10 g m2 Question 6 part 1 of 1 10 points
6
Consider two masses, 4.9 kg and 6.9 kg, connected by a string passing over a pulley having a moment of inertia 10 g m2 about its axis of rotation, as in the figure below. The string does not slip on the pulley, and the system is released from rest. The radius of the pulley is 0.32 m. The acceleration of gravity is 9.8 m/s2 .
0.32 m
Find the linear speed of the masses after m2 descends through a distance 50 cm. Assume mechanical energy is conserved during the motion. 1. 1.03226 m/s 2. 1.06604 m/s 3. 1.09971 m/s 4. 1.13456 m/s 5. 1.17002 m/s 6. 1.20702 m/s 7. 1.24442 m/s 8. 1.2835 m/s correct 9. 1.32335 m/s 10. 1.3643 m/s Explanation:
Let : I = 10 g m2 , R = 0.32 m , m1 = 4.9 kg , m2 = 6.9 kg , h = 50 cm ,
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm v = R, 1 I = M R2 , and 2 1 1 Kdisk = I 2 = M v 2 . 2 4 Consider the free body diagrams T2 T1
7
The assumption that the mechanical energy is conserved and the system starts from rest implies that Kf = U i - U f
6.9 kg m2 g
4.9 kg m1 g
Kf = K1 + K2 + Kpulley 1 = I w2 2 1 v 2 = I 2 r
a
If we neglect friction in the system, the mechanical energy is constant and we can state that the increase in kinetic energy equals the decrease in potential energy. Since Ki = 0 (the system is initially at rest), we have K = Kf - Ki 1 1 1 = m1 v 2 + m2 v 2 + I 2 , 2 2 2 where m1 = 4.9 kg and m2 = 6.9 kg have a common speed. Because v = R, this expression becomes K = 1 2 m1 + m 2 + I R2 v2 .
a
Ui - Uf = Ui1 - Uf 1 + Ui2 - Uf 2 = -m1 g h + m2 g h , this leads to the sam expression for the velocity v as given above. Question 7 part 1 of 1 10 points A child of mass 48.9 kg sits on the edge of a merry-go-round with radius 2.4 m and moment of inertia 222.515 kg m2 . The merrygo-round rotates with an angular velocity of 2.1 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 1.08 m from the center. Now what is the angular velocity of the merry-go-round? 1. 3.33941 rad/s 2. 3.44348 rad/s 3. 3.55442 rad/s 4. 3.66922 rad/s 5. 3.78741 rad/s correct 6. 3.90688 rad/s 7. 4.03125 rad/s 8. 4.15769 rad/s 9. 4.28899 rad/s 10. 4.42429 rad/s Explanation: When the child moves inward, the moment of inertia of the system iM GR + ichild (the merrygo-round plus the child) changes. Therefore, to conserve angular momentum, the angular
From the figure, we see that m2 loses potential energy while m1 gains potential energy. That is, U2 = -m2 gh and U1 = m1 gh. Applying the principle of conservation of energy in the form K + U1 + U2 = 0 gives v= 2 (m2 - m1 ) g h I m1 + m 2 + 2 R = 1.2835 m/s .
Comment: One could work this problem in a systematic way. Begin with the work-energy theorem:
dissip ext Wi->f = Kf - Ki + Uf - Ui + Wi->f
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm velocity of the system must change. Specifically: Linit = Lf inal (ichild + iM GR ) = (ichild,2 + iM GR ) 2 The moment of inertia of the child is m r . Therefore m r 2 + iM GR 2 = . m r2 2 + iM GR Question 8 part 1 of 1 10 points A 23.6 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.45 . The angle between the horizontal and the ladder is 55 . The acceleration of gravity is 9.8 m/s2 .
2
8
9. 1.79947 m correct 10. 1.85799 m Explanation: Let : = 55 , L = 2.8 m , W = m g = 231.28 N , = 0.45 .
and
Nw
f
P ivot
Nf 2.8 m Using the above equilibrium conditions, we have (taking the sum of the torques at the bottom of the ladder) Fx : f - N w = 0 , Fy : Nf - W = 0 , and (1) (2) =0
55 d 23.6 kg
= 0.45 Note: Figure is not to scale. When the person attempts to climb the ladder, how far up the ladder d will the person reach before the ladder slips (kaboom)? 1. 1.39408 m 2. 1.44065 m 3. 1.48731 m 4. 1.53459 m 5. 1.58313 m 6. 1.63536 m 7. 1.68801 m 8. 1.74377 m
: W d cos - Nw L sin = 0 , (3) where d is the distance of the person from the bottom of the ladder. Using Eq. 2, Eq. 3 becomes f L sin = W d cos , Wd so f= . L tan
The ladder may slip when f = fmax = Nw , from Eq. 4 fmax W Wd . L tan
mg
(4)
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm Thus, solving for d , we have 5. T = 2 d L tan (0.45) (2.8 m) tan 55 1.79947 m . (5) 6. T = 2 7. T = 2 8. T = 2 9. T = 2 10. T = 2 7 L correct 12 g 19 L 24 g 14 L 15 g 26 L 21 g 14 L 9 g 62 L 105 g
9
Question 9 part 1 of 1 10 points Hint: The moment of inertia of a uniform 1 rod about its center-of-mass is M L2 . 12 Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal 3 bearing at a point O L from the lower 4 end , as shown in the figure.
Explanation: Basic Concepts: Using the parallel axis theorem, the momentum of inertia is I M d2 . In this case there are two masses with IO = Icm + M D 2 , where 3 1 1 D= - L = L , so 4 2 4 IO = M = 1 + 12 1 4
2
L
O
3 L 4
(1)
L2 (2)
7 M L2 . 48
The period of this pendulum in the small angle approximation is given by 1. T = 2 2. T = 2 3. T = 2 4. T = 2 38 L 63 g 26 L 45 g 2 L 3 g 43 L 72 g
When the rod is at an angle with respect to the vertical direction, the lever arm due to the weight about point O is D sin , which in the small angle approximation is D . In turn the torque due to the weight about point O is given by the equation of motion = I , we have d2 IO 2 = -m g D . dt D is given in Eq. 1 and IO is given in Eq. 2, and for simple-harmonic-motion, we have d2 = - 2 dt2 M gD =- IO
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm 1 M gL =- 4 7 2 ML 48 12 g (3) . = - 7 L The expression for simple harmonic motion and Eqs. 1 and 2, yields 2M g D 2 = , so IO 2 T = 2 IO 2M gD 7 M L2 48 1 M gL 4 7 L . 12 g (4)
10
The total angle rotated in time t = 5 s is given by = 0 t + 1 2 t 2
= (86 rad/s) (5 s) + = 590 rad .
1 (12.8 rad/s2 ) (5 s)2 2
Finally, the number of revolutions rotated is simply = 93.9014 . N= 2 Question 11 part 1 of 1 10 points A particle executes simple harmonic motion with an amplitude of 4.47 cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed? 1. 2.90119 cm 2. 2.99645 cm 3. 3.09171 cm 4. 3.19563 cm 5. 3.29956 cm 6. 3.40348 cm 7. 3.51606 cm 8. 3.63731 cm 9. 3.74989 cm 10. 3.87113 cm correct Explanation: The potential energy of a simple harmonic oscillator at displacement x from the equilibrium point is Uosc = 1 1 k x 2 = m 2 x2 , 2 2
= 2
= 2
Question 10 part 1 of 1 10 points A wheel rotates about a fixed axis with an initial angular velocity of 86 rad/s. During a 5 s interval the angular velocity increases to 150 rad/s. Assume: The angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval? 1. 74.962 2. 77.3493 3. 79.9754 4. 82.6014 5. 85.3071 6. 88.0923 7. 91.0366 8. 93.9014 correct 9. 97.4028 10. 100.666 Explanation: First find f - i = 12.8 rad/s2 . = t
since k = m 2 . When the particle is at maximum displacement A, the energy is all potential: 1 U = m 2 A2 . 2 At other points x, the energy might be both kinetic (speed v) and potential K +U = 1 1 m v 2 + m 2 x2 . 2 2
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm Conservation of energy gives 1 1 1 m v 2 + m 2 x2 = m 2 A 2 , 2 2 2 or v 2 + 2 x2 = 2 A 2 .
11
The speed v is v = -A sin( t) , and the sine is never more than 1, meaning vmax = A . We are asked for the displacement at half this speed, A , v= 2 so conservation of energy is now A 2 or
2
How far below the interface between the two liquids is the bottom of the block? 1. 1.37143 cm 2. 1.42225 cm 3. 1.47822 cm 4. 1.525 cm 5. 1.5836 cm 6. 1.634 cm 7. 1.68543 cm correct 8. 1.73861 cm 9. 1.79351 cm 10. 1.84911 cm Explanation: Given : oil = 930 kg/m3 , h = 3.47 cm , bl = 964 kg/m3 , and water = 1.00 103 kg/m3 . h-y oil
+ 2 x2 = 2 A 2 , WOOD h
1 2 A + x2 = A 2 , 4 from which we see 3 x= A 2 3 (4.47 cm) = 2 = 3.87113 cm . Question 12 part 1 of 1 10 points Oil having a density of 930 kg/m floats on water. A rectangular block of wood 3.47 cm high and with a density of 964 kg/m3 floats partly in the oil and partly in the water. The oil completely covers the block. 3.47 cm oil
3
y
water m , so the volume of the bar is Ay , and v its weight is Wbar = m g = wood (A h) g . Since the block floats, the total buoyant force equals weight of the block, and = Btotal = Wbar Boil + Bwater = Wbar oil [A(h - y)] g + water (A y) g = wood (A h) g oil (h - y) + water y = wood h y (water - oil ) = h (wood - oil ) y=h wood - oil water - oil = (3.47 cm) 964 kg/m3 - 930 kg/m3 1000 kg/m3 - 930 kg/m3
y
WOOD
water
= 1.68543 cm .
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm Question 13 part 1 of 1 10 points There is a moon orbiting an Earth-like planet. The mass of the moon is 3.421022 kg, the center-to-center separation of the planet and the moon is 534000 km, the orbital period of the moon is 27.5 days, and the radius of the moon is 1760 km. What is the angular momentum of the moon about the planet? 1. 2.11907 1034 kg m2 /s 2. 2.19983 1034 kg m2 /s 3. 2.27402 1034 kg m2 /s 4. 2.41028 1034 kg m2 /s 5. 2.49423 1034 kg m2 /s 6. 2.57894 1034 kg m2 /s correct 7. 2.66734 1034 kg m2 /s 8. 2.77022 1034 kg m2 /s 9. 2.85784 1034 kg m2 /s 10. 2.9471 1034 kg m2 /s 1. 8/3 correct 2. 4/3 3. 3/8 4. 3/4 5. 2 6. 2/3 7. 1/2 8. 3/2 Explanation: In the case of axis through the mass m, Im = (2m)L2
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In the case of axis through center of the rod, L 3 L Ic = m( )2 + 2m( )2 = mL2 2 2 4 So Im 8 2mL2 = = 3 2 Ic 3 4 mL Question 15 part 1 of 1 10 points A U-tube of constant cross-sectional area, open to the atmosphere, is partially filled with a heavy liquid with density 9.03 g/cm3 . A light liquid with density 2.49 g/cm3 is then poured into both arms.
Explanation: The angular speed in radians per unit time, for a complete circle is = 2 . T 2 m r2 T kg)
2
Angular momentum L is L = m v r = m r2 = = 2 (3.42 1022
1000 m 1 km 24 hr (27.5 days) 1 day 34 = 2.57894 10 kg m2 /s . (534000 km) Question 14 part 1 of 1 10 points
h light liquid 2.49 g/cm3 0.561 cm heavy liquid 9.03 g/cm3
A massless rod of length L has a mass m fastened at one end and a mass 2m fastened at the other end. What is the ratio of the moment of inertia about an axis through the mass m to the moment of inertia through the center of the rod?
midterm 03 RAMSEY, TAYLOR Due: Nov 15 2006, 8:00 pm If the equilibrium configuration of the tube is as shown in figure, with a difference in the height of the heavy liquid of 0.561 cm, determine the value of the difference in height of the light liquid h . 1. 1.42836 cm 2. 1.47347 cm correct 3. 1.57131 cm 4. 1.63066 cm 5. 1.69958 cm 6. 1.75893 cm 7. 1.81333 cm 8. 1.87276 cm 9. 1.93306 cm 10. 1.99672 cm Explanation: Let : = 2.49 g/cm3 , h = 9.03 g/cm3 , hh = 0.561 cm ,
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and
Also, let h be the height of the light liquid column added to the right side of the U-tube. Consider the pressure at the elevation of the light-heavy liquid interface in the left column and at point at the same elevation in the right column. By Pascal's Principle, the absolute pressure is the same as that at elevation in both columns. But, Pright = Patm + h g hh + g h and Plef t = Patm + g (hh + h + h ) . Thus, from Pascal's Principle, Plef t = Pright (hh + h ) = h (hh ) , or h - 1 hh h = 9.03 g/cm3 -1 = 2.49 g/cm3 = 1.47347 cm .
hh