(1) a = v2/r = 5002 m2/s2/6×103 m = 41.7 m/s2
No. of g's is 41.7/9.8 = 4.25 g. The pilot would feel this plus
an additional 1 g of his weight.
(3) v = 2pr/T = 6.283×1.5×1011 m/3.15×107 s = 2.99×104 m/s
|ar| = v2/r = (2.99×104 m/s)2/1.5×1011m = 6×10-3 m/s2.
The force is F = m ar = 5.98×1024×6×10-3 N = 3.6×1022 N.
The force is exerted by the sun; i.e., the universal gravitational
attraction between the earth and sun.
(5) The tension force required to maintain circular motion of M
is provided by the string. Being constant throughout the length of
the string, this tension is just m g (seen by considering a free
body diagram of the hanging mass). The centripetal acceleration is
a = v2/R, which by Newton's second law applied to
mass M gives F = T = m g = Mv2/R, from which the speed of the
puck is determined to be v = (m G R/M)1/2
(9) A general solution is here provided (for any bank angle q,
including the stated case of q = 0, corresponding to a level
curve).
Letting the x-axis be toward the center of the curve, and y perpendicular to x,
the x and y components of the forces yield
n sinq + nms cosq = m v2/R and
n cosq - nms sinq - m g = 0
(for an assumed friction force acting ``inward''.)
Solving for the coefficient from this pair of equations yields
ms = [v2 cosq/(Rg) - sinq]/[cosq +v2 sinq/(Rg)]
For the indicated parameters, v2/(Rg) = 0.84, which for q = 0
is the answer to the stated problem.
(15) Since the units of v are m/s, and the unit of r is m,
the units of the centripetal acceleration are (m/s)2/m = m/s2.
(17) From the expression in (9) above, one obtains tanq = v2/(Rg)
when ms = 0. From this, the design speed of the curve is computed
to be v = (R g tanq)1/2 = 12.1 m/s. Since the car's speed is
greater at 25 m/s, a friction force acting on the car toward the center
of the curve is necessary. The expression in (9) can be used to solve
for the friction force. Using v = 25 m/s, the
parameter v2/(Rg) = 0.911, which when substituted
into the equation yields ms = [0.911 cos(12°)-sin(12°)]/[cos(12°)+0.911 sin(12°)] = 0.585.
From considerations of (9), the normal force is given
by n = mg/(cosq-ms sinq)
which yields n = 1.37×104 N. Thus the friction force is
f = ms n = 0.585×1.37×104 N = 8×103 N
(25) F = GMm/r2 = GMm/(R+alt)2 = [6.67×10-11×5.98×1024×1400]/[6.37×106 + 1.28×107]2 N = 1.52×103 N
Alternatively, F = m g/9 = 1400×9.8/9 = 1.5×103
from the inverse square law of universal gravity;
i.e., distance is tripled.
(27)g¢ = GM/R2, g = GME/RE2, so g¢/g = MRE2/(ME R2) = 1/2.52 = 0.16, giving g¢ = 0.16 g = 1.6 m/s2
(35) Let the masses be located as follows: one at the origin, one at
x = 0.6 m, y = 0, one at x = 0.6 m, y = 0.6 m, and one at
x = 0, y = 0.6 m. The net force acting on the mass at the origin due to the
other three involves the vector sum of three forces-one strictly along
x, F1, one strictly along y, F2, and the last
one toward the center of the square formed by the masses
F3. Each is of the form F = Gm2/r2 where r = 0.6 m for both
F1 and F2, and r = 0.8485 m for F3. Adding the components
yields Fx = F1 + F3,x = 1.04×10-8N + 3.68×10-9N = 1.41×10-8N = Fy. The magnitude of the net force is
F = (Fx2 + Fy2)1/2 = 2.0×10-8 N and the direction of the
net force is toward the center of the square.
(39) v = (GM/r)1/2 = [(6.67×10-11*5.98×1024/(6.37×106 + 3.6×106)]1/2 m/s = 6.3×103 m/s
(41) a = v2/R = 4.9 m/s2 = (2pR/T)2/R which yields
T = (4p2R/4.9)1/2 = 11 s
(43) T = 2p[r3/(GM)]1/2 = 6.283×[(1.74×106+1×105)3/(6.67×10-11×7.35×1022]1/2 s = 7.1×103 s = 2.0 h.
(45) For the inner radius r1 = 7.3×107 m; likewise the outer is
r2 = 1.7×108 m.
Thus T1 = 6.2832 [(7.3×107)3/(6.67×10-11×5.69×1026)]1/2 = 5.6 h and similarly, T2 = 19.9 h. Alternatively, since the ratio
of r2 to r1 is 2.33, T2 must be greater by 2.333/2.
(67) Will happen for Mm/rm2 = Me/re2 where
re + rm = r, the distance between earth and moon centers; i.e.,
r = 3.84×108 m. These yield a quadratic equation in re:
(Me-Mm)re2 - 2Me r re + Me r2 = 0. Solving by the quadratic
formula gives re = r [(me +/- (Me Mm)1/2)/(Me - Mm)]
Substituting the indicated parameters yields
r = 3.46×108 m (using the negative sign).
(69) Wapp = m (g + a) = m g + m a, so a = 9.8 (80 - 60)/60 m/s2 = 3.3 m/s2.
File translated from TEX by TTH, version 1.95.
On 20 Sep 1999, 12:50.