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Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 5
CHAPTER 5
8.
At each position we take the positive direction in the direction of
the acceleration.
(a) At the top of the path, the tension and the weight are downward.
We write ·F = ma from the force diagram for the ball:
FT1 + mg = mv2/R;
FT1 + (0.300 kg)(9.80 m/s2) = (0.300 kg)(4.15 m/s)2/(0.850 m),
which gives
FT1 = 3.14 N.
(b) At the bottom of the path, the tension is upward and the
weight is downward. We write ·F = ma from the force diagram
for the ball:
FT2 – mg = mv2/R;
FT2 – (0.300 kg)(9.80 m/s2) = (0.300 kg)(4.15 m/s)2/(0.850 m),
which gives
FT2 = 9.02 N.
10. The horizontal force on the astronaut produces the centripetal acceleration:
F = maR = mv2/r;
(7.75)(2.0 kg)(9.80 m/s2) = (2.0 kg)v2/(10.0 m), which gives v =
27.6 m/s.
The rotation rate is
Rate = v/2¹r = (27.6 m/s)/2¹(10.0 m) =
0.439 rev/s.
Note that the results are independent of mass, and thus are the same for all astronauts.
18. The velocity of the people is
v = 2¹R/T = 2¹Rf = 2¹(5.0 m/rev)(0.50 rev/s) = 15.7 m/s.
R
The force that prevents slipping is an upward friction force.
The normal force provides the centripetal acceleration.
We write ·F = ma from the force diagram for the person:
Ffr
x-component: FN = mv2/R;
y
FN
y-component: Ffr – mg = 0.
Because the friction is static, we have
x
mg
Ffr ² sFN , or mg ² smv2/R.
Thus we have
s ³ gR/v2 = (9.80 m/s2)(5.0 m)/(15.7 m/s)2 =
0.20.
There is no force pressing the people against the wall. They feel the normal force and thus are applying
the reaction to this, which is an outward force on the wall. There is no horizontal force on the people
except the normal force.
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Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 5
20. We convert the speeds:
(70 km/h)/(3.6 ks/h) = 19.4 m/s;
FN
y
(90 km/h)/(3.6 ks/h) = 25.0 m/s.
aR
At the speed for which the curve is banked perfectly,
there is no need for a friction force. We take the x-axis
in the direction of the centripetal acceleration.
x
Ffr
We write ·F = ma from the force diagram for the car:
x-component: FN1 sin  = ma1 = mv12/R;

y-component: FN1 cos  – mg = 0.

Combining these, we get v12 = gR tan .
mg
(19.4 m/s)2 (80 m)(9.80 m/s2)tan , which gives
tan  = 0.482, or
 = 25.7°.
At a higher speed, there is need for a friction force, which will be down the incline. If the automobile does
not skid, the friction is static, with Ffr ² sFN.
We write ·F = ma from the force diagram for the car:
x-component: FN2 sin  + Ffr cos  = ma2 = mv22/R;
y-component:
FN2 cos  – Ffr sin  – mg = 0.
We eliminate Ffr by multiplying the x-equation by sin , the y-equation by cos , and adding:
FN2 = m{[(v22/R ) sin  ] + g cos }.
By reversing the trig multipliers and subtracting, we eliminate FN2 to get
Ffr = m{[(v22/R ) cos  ] – g sin }.
If the automobile does not skid, the friction is static, with Ffr ² sFN:
m{[(v22/R ) cos  ] – g sin } ² sm{[(v22/R ) sin  ] + g cos }, or
s ³ {[(v22/R ) cos  ] – g sin }/{[(v22/R ) sin  ] + g cos } = [(v22/gR )] – tan ]/{[(v22/gR ) tan  ] + 1}.
When we express tan  in terms of the design speed, we get
s ³ [(v22/gR ) – (v12/gR )]/{[(v22/gR )(v12/gR )] + 1} = (1/gR )(v22 – v12)/[(v1v2/gR )2 + 1]
= [1/(9.80 m/s2)(80 m)][(25.0)2 – (19.4 m/s)2]/{[(19.4 m/s)(25.0 m/s)/(9.80 m/s2)(80 m)]2 + 1}
=
0.23.
42. We relate the speed to the period of revolution from
v = 2¹R/T.
For the required centripetal acceleration, we have
aR = v2/R = (2¹R/T)2/R = 4¹2R/T2;
9.80 m/s2 = 4¹2(6.38106 m)/T2, which gives T =
5.07103 s (1.41 h).
The result is
independent of the mass
of the satellite.
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