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Unformatted text preview: Mutiple Choice: select the best answer; no partial credit. 1. If the torque on a rigid body is 6.00Nm and the moment arm is 2.00m, what is the
magnitude of the force causing the rotation? ' a.) 1.0N 7;: Fa : (300 NM
b. 6.0N
3.0N F : 6,00Uw _ b.00NM
d.) 2.0N /“‘ ' 2
e.) 10.0N d 00‘“ \ k:3ooM 2. The moment of inertia of a circular disk of radius R and mass M about an axis of
rotation through the center of the disk is I = V2 MR2. What is the moment of
inertial about an axis of rdtation on the edge of disk, parallel to the axis of rotation 
through the Center? (hint: use the parallel axis theorem). D: R 1044: ‘EMR
Ia ~g—MR2.+ MRZ
ligzMR’Z 3. A rigid body rotates with constant angular acceleration a=10.0 rad/s2. If the body
starts at rest, what time does it take to reach an angular speed of (0:200 rad/s? ' w‘c : wc +o<JC
c)3.00S
22:33: w: o
W = oLJC
% ﬂ, t: 0L — [00 vctcL’sZ Problem 4 Two masses hang from a frictionless, massless pulley. Assume the strings have
negligible mass. The values of the masses are m1=2.00kg and m2=3.00kg. C '7 m J A
T T
a.) Draw the free body diagram of each mass.
«4;
A
J i .
F3 ‘3 b.) Find the value of the acceleration of the system a and the tension of the string T.
Use g=10.0 m/sz. (hint: use Newton’s Second Law). ZFY : ¢ A
q \—\M\C} ~ “4ch
#Z\ TAM} a v4ch
(ml—mm : (MIHMQK
«= Mm : amt/n)
M (4+ Wilt“, 3 3.00ltj ’r‘2.oo\:j ‘ S
. Z ,
<2oow><www> Problem 4 (continued) c) Starting from rest, the masses move a distance Ay in t=0.10()s. Find the
work done by the tension T of the string on mass #1. (hint: you must find
Ay which is the displacement in the ydirection). ge‘r A7. y4 = y]. + vii + chLJCL
Vtwo
Q7 ‘ #73 = éatz $3: Jz‘dczj‘
m: ’ if A}: fizwwﬁ
W= (ﬂout). (Jia 1% d Problem 5 Ball #1 initially moVing at speed v, in the positive x direction undergoes a glancing
collision with ball #2, initially at rest. The balls leave at the same angle relative to the x / axis but at different speeds v1f=10.Om/s and V2F5.00m/S. @ V = ‘OOyl
. / \F 5
4.00m, x”
G Vc I ’ ; _._..._.....;;_’ ,.?;_.,_.
\ ©l I M
BEF
ORE AFTER . \ V :5,00MA)
2 S
a.) If m2=1.00kg, use conservation of linear momentum in the y direction to find ml,
the mass of ball #1.
P = P
’t 7%
7.: MV ‘l' MV
0  ‘7f 2 t ' :  (‘01
V‘ : V\ Slqa V17 V243). 9
7% 1° t
V ‘ a mi/ 5‘ 9
:N " “A
O \ \fSu/t 11C .
‘ 1M
M‘V‘ 1F V21; 1
WM: V21: M2 _: S'OOMi‘ (Loom
VH; 10.0 “4' S Problem 5 (continued) b.) Using your result from part a) and conservation of linear momentum in the x
direction, what is the value of vi, the initial velocity of mass #1? F 7
,2. V7—
IMVL l/‘/‘\\:.)(_E + W11 )9? \F
?
V‘M/ 1: \M\V\ Lesa + Mlvzi’cosg
\ L P
C0
‘9 VC 3 3—6 W‘VIF + WZVZ'P)
WI
(059 = id. ~ 35‘
‘3? New ‘+ 'Ma><$'”°“/sﬂ Problem 6 A ball is dropped with mass of 1.00kg with initial downward velocity vi=10.0m/s from a
height 13.0m from the ground. The ball lands on a platform attached to a spring. The
spring’s original length is 3.00m. The ball compresses the spring and comes to rest
1.00m from the ground. What is the spring constant k of the spring? Use g=lO.0rn/s2 and
assume that the gravitational potential energy of the ball is zero at ground level. Neglect
the mass of the platform on the spring. ...
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This note was uploaded on 04/29/2008 for the course PHYS 122 taught by Professor Pope during the Spring '08 term at Clemson.
 Spring '08
 pope

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