Download PHYS 2421 EXAM #5 Wednesday, November 11

PHYS 2421 EXAM #5
Wednesday, November 11, 2015 NAME_______________________
Work three (3) of the following problems from Chapter 12:
43. A horizontal aluminum rod 4.8 cm in diameter projects 5.3 cm from a wall. A 1200 kg
object is suspended from the end of the rod. The shear modulus of aluminum is 32.0 X
1010 N/m2. Neglecting the rod’s mass, find (a) shear stress on the rod and (b) the vertical
deflection of the end of the rod.
𝐹
Δ𝑋
Hints: 𝐴 = 𝐺 𝐿
0
55. In Fig. 12-64, block A (mass 10 kg) is in equilibrium, but it would slip if block KB (mass =
5 kg) were any heavier. For angle  = 30 degrees, what is the coefficient of static friction
between block A, and the surface below it.
𝑇
𝑚 𝑔
Hints𝑇𝐶 cos 𝜃 = 𝑇𝐵 . 𝑎𝑛𝑑 𝑇𝐶 sin 𝜃 = 𝑇𝐴 , tan 𝑇𝐴 = 𝜇 𝑚𝐴 𝑔 , 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝜇𝑠
𝐵
𝑏
67. A solid copper cube has an edge length of 85.5 cm. How much stress must be applied to
the cube to reduce the edge length to 85.0 cm? The bulk modulus of copper is 1.4 X
1011N/m2.
𝐹
Hints: 𝐴 = 𝐵
Δ𝑉 ∆𝑙3
𝑉0
,𝑙
0
3
=
(𝑙+∆𝑙)3 − 𝑙3
𝑙3
=
̃
3𝑙2 ∆𝑙
𝑙3
3
∆𝑙
𝑙
,𝑝 =
𝐹
𝐴
80. A cylindrical aluminum rod, with an initial length of 0.8000 m and radius 1000.0 X 10-6
m, is clamped in place at one end and then stretched by a machine pulling parallel to its
length at the other end. Assuming that the rod’s density (mass per unit volume) does
not change, find the force magnitude that is required of the machine to de crease the
radius to 999.9 X 10 -6 m. (the yield strength is not exceeded.)
𝐹
Δ𝑋
Hints: 𝐴 = 𝐺 𝐿
0
88. The leaning Tower of Pisa is 59.1 m high and 7.44 m in diameter. The top of the tower is
displaced 4.01 m from the vertical. Treat the tower as a uniform, circular cylinder. (a)
What additional displacement, measured at the top, would bring the tower to the verge
of toppling? (b) What angle would the tower then make with the vertical?
𝐷
Hints: θ = tan−1 ( ℎ ) , 𝑙𝑚𝑎𝑥 = ℎ sin 𝜃, Δ𝑙 = 𝑙𝑚𝑎𝑥 − 𝑙
PHYS 2421 EXAM #5
Wednesday, November 11, 2015 NAME_______________________
Work three (3) of the following problems from Chapter 13:
57. In a certain binary-star system, each star has the same mass as our Sun, and they revolve
about their center of mass. The distance between them is the same as the distance
between Earth and Sun. What is their period of revolution in years?
𝐺𝑚1 𝑚2
Hints:
= 𝑚1
𝑟2
(𝜋𝑟/𝑇)2
𝑟/2
, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑇, 𝑢𝑠𝑒 𝑀 = 1.99 𝑋 1030 𝑘𝑔, 𝑟 = 1.5 𝑋 1011 𝑚.
72. A typical neutron star may have a mass equal to that of the sun but a radius of only 10 km.
(a) What is the gravitational acceleration at the surface of such a star? (b) How fast would
an object be moving if it fell from rest through a distance of 1.0 m on such a start? (Assume
the star is not rotating.)
Hints: 𝑎𝑔 =
𝐺𝑀
𝑟2
1
1
1
, 𝐾0 + 𝑈0 = 𝐾 + 𝑈, 𝐾 = 0, 𝐾 = 𝑚𝑣 2 , 𝑣 = √2𝐺𝑀 [ − ]
2
𝑟
𝑟
0
84. Consider a pulsar, a collapsed star of extremely high density, with a mass M equal to that
of the Sun (1.98 X 1030 kg), a radius R of only 12 km, and a rotational period T of 0.041 s. By
what percentage does the free-fall acceleration g differ from the gravitational acceleration
ag at the equator of this spherical star?
Hints: 𝑎𝑔 − 𝑔 = 𝜔2 𝑅, 𝑤ℎ𝑒𝑟𝑒 𝜔 =
2𝜋
𝑇
, 𝑡ℎ𝑒𝑛 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑎𝑔 , 𝑎𝑛𝑑 𝑓𝑖𝑛𝑑 % 𝑑𝑖𝑓𝑓
𝜔2 𝑅
𝑎𝑔
.
90. A 50 kg satellite circles planet Cruton every 6.0 hr. The magnitude of the gravitational force
exerted on the satellite by Cruton is 80 N. (a) What is the radius of the orbit? (b) What is
the kinetic energy of the satellite? (c) What is the mass of planet Cruton?
𝐹
Hints: 𝑚 =
𝑣2
𝑟
, where 𝑣 =
2𝜋𝑟
𝑇
1
, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑟. (𝑏)𝐾 = 2 𝑚𝑣 2 , (𝑐)𝑎𝑔 = 𝐺𝑀/𝑟 2
98. To alleviate the traffic congestion between two cities such as Boston and Washington, D.C.,
engineers have proposed building a rail tunnel along a chord line connecting the cities (Fig. 13-55).
A train, unpropelled by any engine and starting from rest, would fall through the first half of the
tunnel and then move up the second half. Assuming Earth is a uniform sphere and ignoring air drag
and friction, find the city-to-city travel time.
Hints:𝐹𝑔 =
−𝐺𝑀𝑚
3
𝑅
−𝜔2 𝑥, 𝑙𝑒𝑡 𝑇 =
𝑟, 𝐹𝑥 = 𝐹𝑔 sin 𝜃 = [
2𝜋
. 𝑤ℎ𝑒𝑟𝑒
𝜔
𝑇
−𝐺𝑀𝑚
Δ𝑡 = 2 =
3
𝑅
𝜋
.
𝜔
𝑥
𝑟] 𝑟 =
−𝐺𝑀𝑚
3
𝑅
𝑥 𝑜𝑟 𝑎𝑥 =
𝑑2 𝑥
−𝐺𝑀𝑚
− 3 𝑥
𝑑𝑡2
𝑅
=