Classical Mechanics - Final Exam Solution - Physics, Exams of Physics

Prof Walter Lewin, Massachusetts Institute of Technology (MIT) (MA), Physics, Classical Mechanics, Final Exam Solution, cylindrical axis, gravitational acceleration, Friction, maximum displacement angle, period of oscillation.

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2010/2011
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Massachusetts
Institute
of
Technology
-Physics
Department
Physics
- 8
.0
1
Final
Fall
1999
SOLUTIONS
Problem
1
16
points
v
sin
0
4
pts
a)
v
y
=
v
sin
;
gt
=
0
!
t
=
g
0
2
4
pts
b)
y
=
v
+ (
v
sin
)
t
;
1
gt
0 0
2
2
v
sin
1
v
0
sin
0
y
= 0 +
(
v
sin
)
g
;
2
g
0
g
2
y
=
1
v
sin
2
0
2
g
4
pts
c)
v
0
v
sin
0
4
pts
d)
x
=
v
+ (
v
cos
)
t t
=
2
g
0 0
2
v
2
sin
cos
0
x
=
g
1
pf3
pf4
pf5
pf8
pf9
pfa
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Massachusetts Institute of Technology  Physics Department Physics   Final Fall  SOLUTIONS Problem   p oints v 0 sin  pts a vy  v sin  g t   t  (^0) g  pts b y  v   v sin  t (^0 0) g^ t

v 0 sin  (^) v 0 sin  y     v 0 sin  (^) g (^)  g g 2 y   v^ sin (^2)  0  g pts c v 0 v 0 sin  pts d x  v 0  v 0 cos  t t  (^) g v 20 sin  cos  x  g 

Problem   p oints a mg l  mv A^  vA  l

p

 pts g 2  pts b Net force up must b e macen  m  mg T mg  mg T  mg  pts c gravity mg l tension  vA l 

Problem   p oints

pts a p  jrp  F j  bM g sin 

R  pts b Ip  Ic  M b    M M b  

P

pts c p  Ip  bM g sin    ^ M R^   M b^   bg   (^1) R (^2) b 2 sin    2 pts d Under the small angle approximation sin     and the equation bg of motion is given by ^  (^1) R 2 b^2    which is simple harmonic with

2 r

bg an angular frequency given by   (^1) R 2 b 2  The p erio d is given by

s^2

T 

 (^1) R (^2) b 2   2   bg pts e There must b e a force at P or the CM would accelerate straight downwards

Problem p oints pts a The diagram

T

Σ F

mg

pts b       R v   R v    z 2 pts c a  v R a   2 R x  2 pts d F^   ma F^  m   2 R 2 x  pts e T sin   m 2 R  2 T cos   mg tan    2 R g  2

Problem  p oints

r

mv ^  mM^ G^ v  M G pts a  R R v 0 pts b mv R sin  �  mV  R  (^) V   (^0) sin  � ^ 

pts c ^ mv ^ mM^ G

 0 R 2 mM G   m 0 mM^ G  pts d same as in c or 

mV ^ 

v

 R  R 2  pts e ^ mv ^ mM^ G^  ^ m v^0 mM^ G  0 R   2 R  (^)  M G 0 M G

v  

 v

 0 R  ^ R

Problem   p oints pts a M  V 0  pts b W  W M g  Fbuoy Fbuoy  M g W 0  W M g  M g W 0  W

 pts c mg  T  Fbuoy  M g T  M g  mg

W 0

 W  M g Fbuoy  T  W  T



Problem   p oints

m

R

a

2

F

fr

a 1

X X

F  ma   I  no slipping Ffr  ma R Ffr  mR   a a   R Ffr  mR  ma  ma a a

 ^ a a

 ^ a