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Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999
Solutions for Assignment # 10
by Dru Renner [email protected]
December 8, 1999 9:54 am
Problem 10.
(a) We pull with a force F (both ropes combined). At the point O the sidewalk exerts
an unknown force F ′^ on the cylinder. The condition for just lifting the cylinder off the street is N = 0 where N is the usual normal force the street exerts on the cylinder. We can eliminate F ′^ by measuring all torques about O. (A torque is d efined as positive if it would cause the cylinder to roll up the sidewalk.) The condition for just rotating about the corner is τ = 0 where τ is the total torque about O.
0 = F R sin(π − (θ + α)) − M gR sin
( π −
( (^) π
2
− θ
)) = F R sin(θ + α) − M gR cos θ
where we use the trigonometric identities sin(π − x) = sin x andsin
( π 2 +^ x
) = cos x. Solving the above equation for F gives
F
M g
cos θ sin(θ + α)
(b) A local minimum or maximum occurs when
dF dα
= −M g ·
cos θ cos(θ + α) sin^2 (θ + α)
The only solution to the above equation is
cos(θ + α) = 0 =⇒ α = 90◦^ − θ
For θ = 30◦^ then α = 60◦. By plotting (^) M gF we see that α = 60◦^ is a local minimum and the global minimum and α = 0 is the global maximum (not local). It’s not hardto show that if θ < 45 ◦^ then the maximum is at α = 0◦, and if θ > 45 ◦^ then the maximum is at α = 45◦.
Problem 10.
Let L = 5 m, r = 12 · 0 .01 m, A = πr^2 , M = 400 kg, and Y = 0. 36 × 1010 N/m^2 (please see Table 14.1 on page 366 for nylon). Equation (27) on page 366 gives
∆L L
Y
FAPP
A
=⇒ FAPP =
Y A
L
· ∆L
where FAPP is the total appliedforce and∆ L is the corresponding increase in length. It is convenient to define a constant k as
k =
Y A
L
≈ 5. 7 × 104 N/m =⇒ FAPP = k∆L
First we need to calculate the amount the rope is stretched, ∆, due to the weight of 400 kg alone.
M g k
≈ 0 .070 m
(a) We need to find the additional force required to stretch the rope another 0.03 m.
F + M g = k(∆ + 0.03) =⇒ F = k · 0. 03 ≈ 1. 78 × 103 N
(b) Let x denote the displacement down from equilibrium. (Equilibrium corresponds
to the rope length L + ∆.) The equation above gives the necessary appliedforce FAPP to stretch the rope by ∆L; therefore the rope must exert a restoring force of equal and opposite magnitude. Therefore for a displacement of x down from equilibrium, now with the appliedforce not present, the total force is
FTOT = −k(∆ + x) + M g = −kx
where in the above, F is the total force down corresponding to the displacement x down from equilibrium. The above equation is validfor x such that ∆ + x > 0 (x > −∆ ≈ − 0 .07 m) because if ∆ + x < 0 then the rope will develop slack and the restoring force will vanish and the form of the above equation wouldhave to change. For the present, the initial displacement is 0.03 m with no initial velocity, so the motion will never create slack in the rope; hence the above equation remains validfor all the motion. The motion is, therefore, simple harmonic in x with period
T = 2π ·
√ M k
≈ 0 .53 s
(c) With the initial condition of x = 0. 10 m with no initial velocity the motion will
create a slack in the rope. The mass will start at x = 0.10 m proceedthrough x = 0 to
−ρgAx = M x¨
If we use the equivalent expression ρgA = M gd in the above equation andcancel a factor of M we have
g d
· x = ¨x =⇒ T = 2π
√ d g
≈ 3 .5 s
Notice that it was unnecessary to specify the value of kg · m/s 0 to evaluate the expres- sion for T ; hence a change in atmospheric pressure of 5% will not change the periodof oscillation at all.
Problem 10.5 (Ohanian, page 378, problem 63)
Let ρ = 7. 8 × 103 kg be the density of this steel. Now focus on one half of the meter stick, andlet x denote the distance from the axis of rotation to a particular bit of length dx. If the cross-section area is A then the corresponding mass is dm = ρAdx. The centripetal force requiredto accelerate this bit of steel is dF = dm · ω^2 x = ρAω^2 xdx where ω is the angular velocity of the meter stick. The total tension at the center of the meter stick is then
T =
∫ dF =
∫ (^12)
0
ρAω^2 xdx = ρAω^2
x^2
∣∣ ∣∣
1 2 0
ω^2 ρA 8
ω =
√ 8 ρ
T
A
where 12 is the length of half the meter stick. The maximum angular velocity is
ωmax =
√ 8 ρ
( T A
)
max
≈ 624 radian/s
where the ultimate tensile strength for steel,
( T A
) max = 3. 8 × 108 N/m^2 , is given in Table 14.1 on page 366.
Problem 10.6 (Ohanian, page 537, problem 13)
(a) The rate of increase of heat is the amount of energy deposited by each electron
(3. 2 × 10 −^9 J) times the rate for electrons (3. 0 × 1014 s−^1 ).
∆Q ∆t
= 3. 2 × 10 −^9 · 3. 0 × 1014 = 9. 6 × 105 J/s ≈ 230 kcal/s
where the conversion usedis 1 cal = 4 .186 J.
(b) The mass of water in the beam dump is m = 10^3 × kg/m^3 · 12 m^3 = 1. 2 × 104 kg.
Equation (1) on page 517 is
∆Q = mc∆T
Therefore the rate of increase of temperature is
∆T ∆t
mc
∆Q
∆t
≈ 0. 019 ◦C/s
where the specific heat capacity for water is given by Table 20.1 on page 516 as c = 1 .00 kcal/kg ·◦^ C.
Problem 10.7 (Ohanian, page 512, problem 17)
The volume of one mole of gas at STP is
V =
RT
p
≈ 0 .0227 m^3
The volume occupiedby the helium atoms is
Vatom ≈ 6. 02 × 1023 · 3 · 10 −^30 ≈ 1. 81 × 10 −^6 m^3
Therefore the fraction of volume occupiedby the helium atoms is
f =
Vatom V
≈ 8 × 10 −^5 = 0.008%
Problem 10.8 (Ohanian, page 513, problem 24)
(a) Let p = 1 atm = 1. 01 × 105 N/m^2 , A be the cross-section area of the diving bell,
h = 2 m, and h′^ be the height of air in the diving bell once it is immersed. Then p′, the pressure of air in the diving bell once it is immersed, is given by the pressure of the water at a depth of 15 m.
p′^ = p + ρgz
where ρ = 10^3 kg/m^3 is the density of water, and z = 15 m. This gives the value
p′^ = 2. 48 × 105 N/m^2
Assuming the temperature is constant, then
phA = pV = p′V ′^ = p′h′A =⇒ h′^ =
( p p′
) h ≈ 0 .81 m
Problem 10.10 (Ohanian, page 512, problem 21)
The density of the external air is ρa = 1.20 kg/m^3. The mass of the balloon, etc. without the air is Mb = 730 kg, andthe volume of the balloon is Vb = 2200 m^3. If the mass of the hot air inside the balloon is M andthe equivalent mass of coldair (same volume but atmospheric temperature) is ρaVb, then the condition for the balloon to just lift-off is
ρaVb − M = Mb =⇒ M = ρaVb − Mb = 1. 91 × 103 kg
The balloon is open at the bottom, so the pressure of the air within is the same as the pressure of the external air. The mass per mole of air is given in Example 3 on page 498 as m = 29.0 g/mole. The pressure of the external air at Ta = 20◦C = 293.15 K is
pa =
naRTa Va
nam Va
RTa m
= ρa ·
RTa m
= 1. 0086 × 105 N/m^2
The air within the balloon has the mass
M = mnb = m ·
pbVb RTb
=⇒ Tb = m ·
pbVb RM
= 405 K = 132◦C
where pb = pa.
Problem 10.11 (Ohanian, page 513, problem 28)
Consider a slab of air with horizontal area A andheight dh. The pressure difference between the bottom andthe top of the slab must support the weight of the air within the slab.
pA − (p + dp)A − (A dh) · ρ · g = 0 =⇒ dp = −ρg dh (2)
If we wouldassume that ρ is a constant, then we wouldderive the usual result for pressure, but if the temperature remains constant with height, then the density must vary with height. We can obtain a relationship between the pressure and density by using the ideal gas law.
p =
N
V
· kT =
mN V
kT m
= ρ
kT m
=⇒ ρ =
mp kT
where m = (^29) N.0 gA = 4. 82 × 10 −^26 kg is the mass of one molecule of air. We can use the above to write Equation (2) as
dp = −p ·
mg kT
dh
This is now a validdifferential equation for the pressure, p, as a function of height, h. We can solve this equation by following these steps.
p
dp = −
mg kT
dh
ln p − ln p 0 =
∫ (^) dp
p
mg kT
∫ dh = −
mg kT
(h − 0)
p = p 0 e−^
mghkT
Problem 10.12 (Ohanian, page 535, problem 5)
The lateral loops act as gaps to allow for thermal expansion in the long oil pipelines. Even for a small change in temperature, the increase in length of the pipeline couldbe very large. This would prevent the pipeline from being attached rigidly at each end. The loops allow each shorter section of pipe to stretch but prevent the net effect from being cumulative. Since the oil must still flow loops must be usedinsteadof simple gaps as in bridges.
