Massachusetts Institute of Technology
Department of Physics
Physics 8.01L
SAMPLE FINAL EXAM
SOLUTIONS
Problem 1
a) V =
N kT
P
, VN EW =
N k T3
P
10
=
10 N kT
3 P
,
⇒ V goes up
b) (i) F = mg = 8(10) = 80N
(ii)FBuoy = (1000)(.002)(10) = 20N,
∴ F = 80 − 20 = 60N
c) Velocity goes down, pressure goes up.
Problem 2
a) Equilibrium, so FBuoyant − w = 0 ⇒ FB = w.
FB = Vf ρf g = (πr2 d)ρg = w, d = πrw2 ρg
2
b) FT OT = M a = FB − w, FB = πr2 ( d2 )ρg = d( πr2ρg )
2
From (a), d = πrw2 ρg ⇒ FB = πrw2 ρg ( πr2ρg ) = w2
Half as deep ⇒ half as large buoyancy force.
−w
FT OT = M a = w2 − w = −w
2 , a = 2m , but w = M g.
−g
⇒ a = 2 , accelerating downward at g2 .
Problem 3
a) P + ρgy = constant, P1 + ρg(0) = P2 + ρg(6500)
P2 = P1 − ρg(y2 ) = 1.013 × 105 − (0.95)(9.8)(6500),
P2 = 4.08 × 104 N/m2 = 0.40 atm
b) N = const, V = const, P V = N kT
⇒
P1
T1
=
P2
T2 ,
T1 = 293, P1 = 1.5 × 107 , T2 = 253, P2 =
P1 T2
= 1.30 × 107 N/m2
T1
Problem 4
ii) f exerts torque around center of mass, so you fall over.
1
January 25, 2006
iii) Now N exerts torque which can balance torque due to friction.
Problem 5
a)
b) Take torques around toes: M gL cos(θ) − F ( 43L ) cos(θ) = 0, F =
c)T + F = M g, T =
3
Mg
4
1
Mg .
4
Problem 6
a) 3M g + M g = 4M g.
b) Take torques about left end: 4M gD − LM g = 0, D = L4 .
Check torque around weight: 0 = D(3M g) − M g(L − D), D(4M g) = M gL, D =
Problem 7
a)
b)
�
c)
d)
�
Fx = N2 + N1 sin(θ) − f1 cos(θ) = 0
Fy = f2 − M g + N1 cos(θ) + f1 sin(θ) = 0
�
�
τ = M g( L2 cos(θ)) − N1 L = 0.
τ = N2 L sin(θ) + f2 L cos(θ) − M g L2 cos(θ) = 0.
Problem 8
a) τ = Iα, take torques about hinge.
2
(M g)( L2 )(sin(90◦ )) = ( ML
3 )(α) ⇒ α =
gL
2
L2
3
=
3g
2L ,
α=
2
3g
2L
L
4.
b)
F = M acm , acm = α( L2 ), downward.
All forces and acceleration are vertical ⇒ FH = 0 .
3g
)( L2 ) = −3Mg
FV − M g = −M a = −M α( L2 ) = −M ( 2L
4
Mg
Mg
3Mg
FV = M g − 4 ⇒ FV =
, FT OT =
, up .
4
4
c) Used fixed pivot:
KEI = 0, P EI = M gL, KEF = 12 Iend ω 2 , P EF = M g( L2 ), Work = 0.
�
3g
3g
MgL
2
L
ML2 ω 2
1 ML2
2
or ω =
= 2 , ω =
.
2 ( 3 )ω + M g 2 = M gL,
6
L
L
Used center of mass:
2
+ 21 ICM ω 2 , vCM = ω( L2 )
KEI = 0, KEF = 21 M vCM
1
L 2 2
1 ML2
1
3
1
KEF = 2 M ( 2 ) ω + 2 ( 12 )ω 2 = M L2 ω 2 ( 81 + 24
) = M L2 ω 2 ( 24
+ 24
)
2 2 1
= M L ω ( 6 ) ⇒ Same answer.
Problem 9
a) L is conserved: mvR = (I0 + mR2 )ωf
mvR
ωf =
I0 + mR2
2 2
2
1 m v R
2
b) KEI = 12 mv 2 , KEF = 21 (I0 + mR2 )( I0mvR
+mR2 ) = 2 ( I0 +mR2 )
mR2
KEF
=
KEI
I0 + mR2
Problem 10
a) Left
b) Yes, gravity.
c) Out of the page; Counter-clockwise.
d) Yes, pivot force.
e) Out of the page; Counter-clockwise.
f ) Out of the page.
Problem 11
Take clockwise to be positive. Angular momentum is conserved: IωI − mvI d = Iωf + mvf d
0.30(ω) − 0.15(50)(0.8) = 0.3(0.35ω) + 0.15(40)(0.8)
0.20ω = 6 + 4.8 ⇒ ω = 54rad/s . Period = 0.12 sec.
3