Unformatted text preview: g (Equation 4.5), so we can divide the expression for the
centripetal force by the expression for the weight and obtain that
Fc = FN − W
W = mv 2
mg r FN or W −1 = v2
gr Solving for the ratio FN /W, we find that ( 230 m/s )
v2
= 1+
= 1+
= 8.8
W
gr
9.80 m/s 2 ( 690 m ) FN 2 ( ) 45. REASONING The magnitude Fc of the centripetal force is Fc = mv 2 / r (Equation 5.3).
Since the speed v, the mass m, and the radius r are fixed, the magnitude of the centripetal
force is the same at each point on the circle. When the ball is at the three o’clock position,
the force of gravity, acting downward, is perpendicular to the stick and cannot contribute to
the centripetal force. (See Figure 5.21 in the text, point 2, for a similar situation.) At this
point, only the tension of T = 16 N contributes to the centripetal force. Considering that the
centripetal force is the same everywhere, we can conclude that it has a magnitude of 16 N
everywhere.
At the twelve o’clock position the tension T and the force of gravity mg both act downward
(the negative direction) toward the center of the circle, with the result that the centripetal
force at this point is –T – mg. (See Figure 5.21, point 3.) At the six o’clock position the
tension points upward toward the center of the circle, while the force of gravity points
downward, with the result that the centripetal force at this point is T – mg. (See Figure 5.21,
point 1.)
SOLUTION Assuming that upward is the positive direction, we find at the twelve and six
o’clock positions that 270 DYNAMICS OF UNIFORM CIRCULAR MOTION − T − mg = −16 N
1 24
43 Twelve o' clock Centripetal
force b g
c h g
c h T = 16 N − 0.20 kg 9 .80 m / s 2 = 14 N
T − mg = 16 N
123 Six o' clock Centripetal
force b T = 16 N + 0.20 kg 9 .80 m / s 2 = 18 N 46. REASONING Because the crest of the hill is a circular arc, the
motorcycle’s speed v is related to the centripetal force Fc acting FN on the motorcycle: Fc = mv 2 r (Equation 5.3), where m is the
mass of the motorcycle and r is the radius of the circular crest.
Solving Equation 5.3 for the speed, we obtain v 2 = Fc r m or mg v = Fc r m . The freebody diagram shows that two vertical
Freebody diagram of
forces act on the motorcycle. One is the weight mg of the
the motorcycle
motorcycle, which points downward. The other is the normal
force FN exerted by the road. The normal force points directly
opposite the motorcycle’s weight. Note that the motorcycle’s weight must be greater than
the normal force. The reason for this is that the centripetal force is the net force produced by
mg and FN and must point toward the center of the circle, which lies below the motorcycle.
Only if the magnitude mg of the weight exceeds the magnitude FN of the normal force will
the centripetal force point downward. Therefore, we can express the magnitude of the
centripetal force as Fc = mg − FN. With this identity, the relation v = Fc r m becomes v= ( mg − FN ) r
m (1) SOLUTION When the motorcycle rides over the crest sufficiently fast, it loses contact with
the road. At that point, the normal force FN is zero. In that case, Equation (1) yields the
motorcycle’s maximum speed: v= ( mg − 0 ) r
m = m gr
= gr =
m (9.80 m/s2 ) ( 45.0 m) = 21.0 m/s Chapter 5 Problems 271 47. REASONING When the stone is whirled in a horizontal circle, the centripetal force is
provided by the tension Th in the string and is given by Equation 5.3 as
mv 2
Th =
{
r (1) Centripetal
force where m and v are the mass and speed of the stone, and r is the radius of the
circle. When the stone is whirled in a vertical circle, the maximum tension
occurs when the stone is at the lowest point in its path. The freebody
diagram shows the forces that act on the stone in this situation: the tension
Tv in the string and the weight mg of the stone. The centripetal force is the
net force that points toward the center of the circle. Setting the centripetal
force equal to mv 2 / r , as per Equation 5.3, we have
mv 2
+Tv − mg =
14 3
24
r Tv Stone
mg
(2) Centripetal
force Here, we have assumed upward to be the positive direction. We are given that the maximum
tension in the string in the case of vertical motion is 15.0% larger than that in the case of
horizontal motion. We can use this fact, along with Equations 1 and 2, to find the speed of
the stone. Solution Since the maximum tension in the string in the case of vertical motion is 15.0%
larger than that in the horizontal motion, Tv = (1.000 + 0.150) Th . Substituting the values of
Th and Tv from Equations (1) and (2) into this relation gives
Tv = (1.000 + 0.150 ) Th mv 2 mv 2
+ mg = (1.000 + 0.150 ) r
r
Solving this equation for the speed v of the stone yields
v= gr
(9.80 m/s 2 ) (1.10 m)
=
= 8.48 m/s
0.150
0.150 272 DYNAMICS OF UNIFORM CIRCULAR MOTION Clothes 48. REASONING The drawing at the right
shows the two forces that act on a piece of
clothing just before it loses contact with the
wall of the cylinder. At that instant the
centripetal force i...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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