RECITATION PROBLEMS
CHAPTER 6: CIRCULAR MOTION AND GRAVITATION
6-3 Force on a skater’s wrist. A 52 kg ice skater spins about a vertical axis through her
body with her arms horizontally outstretched, making 2.0 turns each second. The
distance from one hand to the other 1.50 m, Biometric measurements indicate that
each hand typically makes up about 1.25% of body weight.
(a) Draw a free-body diagram of one of her hands.
(b) What horizontal force must her wrist exert on her hand?
(c) Express the force in part (b) as a multiple of the weight of her hand.
Free-body diagram
y
arad
F
~
Fy
x
Fx
w
(b) a rad 
4 2 R 4 2 (0.750m)

 118m / s 2
2
2
T
(0.50s)
Fx  marad  (0.65kg)(118m / s 2 )  77 N
(c)
F 77 N

 12 N
w 6.4 N
The horizontal force from the wrist is 12 time the weight of the hand.
1
6-5 The “Giant Swing” at a county fair consists of a vertical central shaft with a number
of horizontal arms attached at its upper end. Each arm supports a seat suspended
from a 5.00-m-long rod, the upper end of which is fastened to the arm at a point 3.00
m from the central shaft.
(a) Make a free-body diagram of the seat, including the person in it.
(b) Find the time of one revolution of the swing if the rod supporting the seat makes
an angle of 30.0° with the vertical.
(c) Does the angle depend on the weight of the passenger for a given rate of
revolution?
(a) Free-body diagram:
In the following figure θ = 30.0°.
y
F
arad
~
Fcos30.0°
θ
x
Fsin30.0°
w
2
(b)
F
y
 ma y gives F cos 30.0  mg  w . Solving this equation for F gives,
mg
------------------ (1)
cos 30.0
F
The person moves in a circle of radius R  3.00m  5.00msin 30.0  5.50m.
The acceleration of the person is a rad 
v2
directed horizontally to the left as
R
shown in the figure.
F
x
 max gives
mv 2
------------------- (2)
R
From equations (1) and (2) we get,
mg
mv 2
sin 30.0 
cos 30.0
R
F sin 30.0 
v  Rg tan 30.0  (5.50m)(9.80m / s 2 ) tan 30.0  5.58m / s
The time for one revolution is the period T 
T
2R
2 5.50m 
.
 6.19 s.
v
5.58m / s
2R
.
v
(c) The angle does not depend on the weight of the passenger for a given rate of


rotation since the net force is proportional to m and in  F  ma when the
expression for F is substituted (see above) the mass cancels out.
3
6-13 Effect on blood of walking. A person is walking, his arms swing through
approximately a 45° angle in (½) s. As a reasonable approximation, we can assume
that the arm moves with constant speed during each swing. A typical arm is 70.0
cm long, measured from the shoulder joint.
(a) What is the acceleration of a 1.0 gram drop of blood in the fingertips at the bottom
of the swing?
(b) Make a free-body diagram of the drop of blood in part (a).
(c) Find the force that the blood vessel must exert on the drop of blood in part (b).
Which way does the force point?
(d) What force would the blood vessel exert if the arm were not swinging?
(a) A 45° angle is
1
1
of a full rotation, so in s a hand travels through a distance of
8
2
1
2R  .
8
The speed and hence the radial acceleration is calculated as follows:
1  2R 
v 
  1.10m / s
8  0.50s 
2
v 2 1.10m / s 
a rad 

 1.73m / s 2
R
0.700m
(b) In the following free-body diagram, F is the force exerted by the blood vessel.
+y
F
arad
+x
w = mg
(c)
F
y
 ma y gives F  w  marad and



F  mg  a rad   1.00  10 3 kg 9.80m / s 2  1.73m / s 2  1.15  10 2 N, upward.
(d) When the arm hangs vertically and is at rest, F  w  marad  0 and hence,


F  w  mg  1.00  10 3 kg 9.80 N   9.80  10 3 N . .
4
6-22 Find the magnitude and direction of the net gravitational force on mass A due to
masses B and C in the following figure. Each mass is 2.00 kg.
(a)
Force on mass A because of mass B is directed horizontally to the right (+x
direction) and is calculated as follows:
FB  G
m A mB
rAB
2

 6.673  10 11 N .m 2 / kg 2
 2.00kg
2
0.50m
2
 1.069  10 9 N
Force on mass A because of mass C is directed horizontally to the right and is
calculated as follows:
FC  G
2.00kg  2.669 108 N
mAmC
 6.673 1011 N .m 2 / kg 2
2
0.10m2
rAC


2
The net force on mass A because of masses B and C is:
Fnet  FB  FC  1.069 10 9 N  2.669 10 8 N  2.8 10 8 N
This force is towards the right.
(b) Force on mass A because of mass B is directed horizontally to the right and is
calculated as follows:
FB  G
m A mB
rAB
2

 6.673  10
11
2
N .m / kg
2
2

2.00kg

0.40m2
 1.668  10 9 N
Force on mass A because of mass C is directed horizontally to the left and is
calculated as follows:
2.00kg  2.669 108 N
m m
FC  G A 2C  6.673 1011 N .m 2 / kg 2
0.10m2
rAC


2
The net force on mass A because of masses B and C is:
Fnet  FB  FC  1.069  10 9 N  2.669  10 8 N  2.5  10 8 N
This force is towards the left.
5
6-24 Each mass in the following figure is 3.00 kg. Find the magnitude and direction of
the net gravitational force on mass A due to the other masses.
Let in figure (a), the force on mass A because of mass B has a magnitude FB and is
pointing towards mass B. Hence this force vector is making an angle o f 30° with
the downward direction. Also, suppose the force on mass A because of mass C is
Fc. This force is pointing towards mass C and is oriented at an angle of 30° with
the downward direction. Magnitudes of these forces can be calculated as follows:
2
m A mB
11
2
2 3.00kg 

6
.
673

10
N
.
m
/
kg
 6.00  10 8 N .
2
2
rAB
0.100m 
x and y components are:

FA  FC  G

Fx  FBx  FCx  0, Since FCx   FBx
Fy  FBy  FCy  2FB cos 30  1.04 10 7 N.
The net force is 1.04 x 10-7 N, in the direction from A toward the center of the line
connecting B and C.
In figure (b), the distance between A and D is rAD = 0.141 m
2
m A mD
11
2
2 3.00 kg 
FD  G
 6.673  10 N .m / kg 
 3.02  10 8 N .
2
2
0.141m 
rAD
FB  FC  G
m A mB
rAB
2

 6.673  10 11 N .m 2 / kg 2
6
 3.00kg 
2
0.100m 
2
 6.00  10 8 N .
Let +x direction be towards the right and +y upwards.
Following are the x and y components of the net force:
Fx  FBx  FCx  FDx  6.00  10 8 N  0  3.02  10 8 N cos 45  8.13  10 8 N .
Fy  FBy  FCy  FDy


Fy  0  6.00  10 8 N  3.02  10 8 N sin 45  8.13  10 8 N .
The magnitude of the net gravitational force on mass A is
F  Fx2  Fy2 
8.13 10
8
N
  8.13 10
2
8
N

2
 1.15  10 7 N .
This force is pointing towards mass D.
6-29 Huygens probe on Titan. In January 2005 the Huygens probe landed on Saturn’s
moon Titan, the only satellite in the solar system having a thick atmosphere.
Titan’s diameter is 5150 km, and its mass is 1.35 x 1023 kg. The probe is weighed
3120 N on the earth. What did it weigh on the surface of Titan?
The mass of the probe can be determined using the expression for its weight on
w
3120 N
earth m  E 
 318.4kg.
g E 9.80m / s 2
1
The radius of Titan is rT  5150km  2.575  10 6 m
2
The weight on the surface of Titan is:
wG
mmT
rT
2



(6.673  10 11 N .m 2 / kg 2 )(318.4kg) 1.35  10 23 kg
 433N
(2.575  10 6 m) 2
Weight depends on the location, mass does not.
6-36 Planets beyond the solar system. On October 15 2001, a planet was discovered
orbiting around the star HD68988. Its orbital distance was measured to be 10.5
million kilometers from the center of the star, and its orbital period was estimated at
6.3 days. What is the mass of HD68988? Express your answer in kilograms and in
terms of our sun’s mass.
Following is the equation for the period T:
T
2r 3 / 2
Gm
OR T 2 
4 2 r 3
Gm
Solving this equation for the mass m, we get mass of the star HD68988 as follows:
m
4 2 r 3
4 2 (10.5  10 9 m) 3

 2.3  10 30 kg
T 2G
(5.443  10 5 s) 2 (6.673  10 11 N .m 2 / kg 2 )
7
Mass of HD68988 in terms of our Sun’s mass is:
2.3  10 30 kg  ms 2.3  10 30 kg  ms
m

 1.2ms
ms
1.99  10 30 kg
8