This preview shows pages 1–8. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1996 Physics C Solutions Distribution of points
Mech. 1 (15 points)
(a) 2 points
' Average Time
for Ten Vibrations
For recording the correct values for the period 1 point
For recording the correct values for the square of the period 4 1 point (b) 3 points The following shows the correctly graphed data and a sample of an TRé—)\ g acceptable lme to ﬁt the data» ~ 7 ﬁr 4.0 _'__'I"__——_I'_—"‘I'""'T _ _ ' ' ' — ‘ _  — — '  "‘I """""" "'I— I"
I I I I I I I
I I  I I I I
I I I I I I
————I——————I——————I————4 — — —  — — — — — — — — — — — a — — — — — — — — I————I
I I I I I I
I I I I I I
I I I I I O
3.0 I——I—+—'r  'r'
I I  I I I
I I I I I 
I I I I I
—____L___L—__.._L___J. _ _ _ _ _ _ _ _ _ _ _ _ _ . _ .._I__...__L...___
I I I I I I I
I I _  I I I I
: : : : ' ' '
2,0 “"‘I‘“‘‘f‘"'T"“‘T ‘ _ _ _ ' ‘ ’ “ ‘ ‘  “ I'"""I"_"‘I
I I I I I I I
I _ I I I I I I
I I I I I I 
__._.L__._...L____J._____J. _ _ . _ . _ _ _ . _ _ . _ _ . _ . . . . _ _ _ . _...__._.._L___._._I
I I I I I I
I I I I I 
I I I I I I
I I I I I I "'"'—'I'—"'T'    ‘  —   ' ~ — —    —   — —   I—"I~'I
I O I
I I
_____l. ______________________________________________ “g. ____ _.
I I
I I
. Mass
0.10 0.20 0.30 0.40 0.50 I)
For correctly plotting the values of T2 1 point
For drawing a straight line that is not horizontal or vertical 1 point For drawing a straight line consistent with the data, having at least one point
above and one point below the line 1 point (C) 2 points The period is T= (16.1 s)/10 = 1.61 5. Thus T2 = 2.59 52. (d) 2 points (6) (f) For a value of mass for this peroid that is consistent with the graph 1 point
On the sample graph given, the corresponding mass as read ﬁ'om the straightline ﬁt is 0.45 kg.
For an answer expressed with 2 or 3 signiﬁcant ﬁgures 1 point
For recognizing this is simple harmonic motion and using the equation for the period of a spring , T = 2n‘/::: 1 point
For solving the equation for k 1 point
k = 41:2 1'2— T The ratio sz— is the inverse of the slope of the straightline ﬁt.
(Alternate solution) (Alternate points)
For attaching one or more masses to the bar and measuring the force and displacement at equilibrium 1 point
For indicating that the force constant can be determined from a graph of force versus displacement or the equation k = F/x 1 point
2 points
For indicating that this device can be used in space 1 point
For explaining that the period of oscillation is independent of the effects of gravity 1 point
3 points
For indicating that the gravitational force is equal to the net force on the shuttle 1 point GmMe
FG = 2 = ma
r
For solving for the acceleration 1 point
GMe
a = r2
r = (6.4 x106 m/s) + (0.3 x106 m/s) = 6.7 x106 m/s
3
a = [6.65 x 10‘” m z](6.0 x 1024 kg)/(6.7 x 106m)2
kg  s For the correct answer 1 point a = 8.9 m/s2 (g) lpoint For explaining that all objects aboard the shuttle are accelerating toward the Earth at the same rate as the shuttle, or thtat they are all in “freefall”. 1 point
Without a contact force counteracting the force of gravity, there is no sensation of weight. 1996 Physics C Solutions Mech. 2 (15 points) (a o) @) (Q 3 points For indicating, via an equation or a ﬂeebody diagram, that the net force
is the sum of gravity and the normal force exerted by the platform Using Newton’s second law mg  N = mav Solving for the force exerted by the platform N=m@m) For correctly substituting both accelerations N: (300 kg)(10 m/s,2  1.5 m/sz) For the correct answer N = 2550 N (or 2490 N if using 9.8 111/52) 2 points For indicating that the ﬁ’ictional force is the only horizontal force exerted f = man
f= (300 kg)(2 m/sz) For the correct answer
f = 600 N 3 points Expressing the frictional force in terms of the normal force
f=NV u =f/N For the correct substitution of both forces u = (600 N)/(2,550 N)
For the correct_ answer, with no units
it = 0.24 4 points
For writing the equation for the vertical motion, and indicating that av = 1.5 y = y., + %(—1.5)r2 For indicating that y0 = 2
For writing the equation for the horizontal motion, and indicating that ah = 2 1
=——2t2
x J) Distribution
of points 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point
1 point 1 point For a correct equation relating y and x For a straight line with a negative sicpea, extending at least to x = 1 m For a yintercept at 2 m For an xintercept at 2 2/3 m or a slope of 3/4 1996 Physics C Solutions
Distribution
of points Mech. 3 (15 points)
(a) 2 points Beginning with the integral expression for rotational inertia I, = Irzdm For a correct expression for the mass element dm 1 point dm=£dr
I For correct limits on the integral 1 point t/2
Ir = rzdr
Z
(/2 (Alternate solution) (Alternate Points) For correctly using the parallel axis theorem 1 point
Iend = 1‘,m 4 mr2 1and = %M£2 Solving for 1“; [em = Iend —mr2 For the correct substitutions I point 2
1m, =1MeZ—M54— (b) 2 points For adding terms for the rod and the hoop to obtain the total rotational inertia I = Ir + Ih
For the correct expressions for each of the terms
2 2
I = M! + M £
12 4
2
1 = M.
3 (c), and (e) 8 POiIltS The solutions to these three parts are highly connected, and so points for the
various principles involved in their solution were awarded independent of the part in which they appeared. For use of Newton’s second law Fm =ma For correctly determining the net force on the cat Fm = ng " 71 Ma = Mg  T For the torque equation for the rotation of the rodhoop assembly 1 = let
For the correct torque exerted by the rope 1:=T£
2 For the correct relationship between the angular and the linear accelerations
0L =a/R or a/(l/Z) Solution for part (0):
Substituting the torque from the rope, the angular acceleration, and the
rotational inertia in the torque equation
T5 = M a
2 3 Z Solving for the acceleration 3 T
a = ——
4 M
Substituting a in the Newton’s law equaiton and solving for T 1 point 1 point 1 point 1 point 1 point 1 point 1 point (f) 3
—T=M —T
4 g 7
—T=M
4 3 For the correct answer 4
T=—M
7 3 Solution for part (d):
Using the torque equation
1' = T—l; = [on 2 “If
21 Substituting for T and I =§ (state) For the correct answer a = E 5
7 6
Solution for part (e): we) Substituting for a “(sat:4) For the correct answer r:
7g
3points For a correct expression for the angular momentum of the cat L=Mu§ or Mur For a kinematic equation that allows calculation of the speed o of the cat oz =u§ —Zad zeta = 10 3
For the correct answer 2 —Mwl%g£ or MJZaH—g 1 point 1 point 1 point 1 point 1 point 1 point ...
View Full
Document
 Spring '09
 Park
 Physics

Click to edit the document details