Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004. 1. This
print-out should ... An amusement park ride consists of a rotating circular platform
...
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. Chapters 5 and 6 problems. 001 (part 1 of 2) 5 points An amusement park ride consists of a rotating circular platform 11.4 m in diameter from which 10 kg seats are suspended at the end of 1.18 m massless chains. When the system rotates, the chains make an angle of 43.7◦ with the vertical. The acceleration of gravity is 9.8 m/s2 .
002 (part 2 of 2) 5 points If a child of mass 62.5 kg sits in a seat, what is the tension in the chain (for the same angle)? Correct answer: 982.756 N. Explanation: M = 62.5 kg From the first part we have T cos θ = (m + M ) g (m + M ) g T = cos θ (10 kg + 62.5 kg) (9.8 m/s2 ) = cos 43.7◦ = 982.756 N .
d l
1
θ
What is the speed of each seat? Correct answer: 7.81126 m/s. Explanation: In the vertical direction we have
003 (part 1 of 1) 0 points A figure of a dancer on a music box moves counterclockwise at constant speed around the path shown below. The path is such that the lengths of its segments, P Q, QR, RS, and SP , are equal. Arcs QR and SP are semicircles. S R P
T cos θ = m g , where T is the tension in the chain. In the horizontal direction we have T sin θ =
m v2 . r
Since r = ` sin θ +
d 2
= (1.18 m) sin 43.7◦ +
Q
Which of the following best represents the magnitude of the dancer’s acceleration as a function of time t during one trip around the path, beginning at point P ? a 1. O tP tQ tR tS tP
11.4 m 2
2.
t
a
= 6.51524 m , we have p v = g r tan θ q = (9.8 m/s2 ) (6.51524 m) tan 43.7◦ = 7.81126 m/s .
O tP tQ tR tS tP
3.
t
a
O tP tQ tR tS tP
t
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004 2 p a 3. v = 2 µ g R 4. s gR t correct 4. v = O tP tQ tR tS tP µ 1p a 5. 5. v = gR µ p 6. v = 2 gR t O tP tQ tR tS tP p 7. v = µ g R correct p Explanation: 8. v = µ 2 π g R During RS and P Q, the acceleration is zero p because the velocity is a constant both in mag9. v = 2 µ g R nitude and in direction. During QR and SP , p even though the magnitude of the velocity is 10. v = g R a constant, it changes its direction, which will Explanation: v2 result in an acceleration of , where v is the Basic Concepts: r m v2 velocity and r is the radius of the arc. So the Centripetal force: F = acceleration is fixed at a non-zero constant r Frictional force: fs ≤ µ N = fsmax during SP and QR. Solution: The maximum frictional force due to friction is fmax = µ N , where N is the 004 (part 1 of 2) 0 points inward directed normal force of the wall of An amusement park ride consists of a large the cylinder on the person. To support the vertical cylinder that spins about its axis person vertically, this maximal friction force fast enough that any person inside is held fsmax must be larger than the force of gravity up against the wall when the floor drops away m g so that the actual force, which is less (see figure). The coefficient of static friction than µN , can take on the value m g in the between the person and the wall is µ and the positive vertical direction. Now, the normal radius of the cylinder is R. v2 force supplies the centripetal acceleration R R on the person, so from Newton’s second law, N =
m v2 . R
Since ω What is the minimum tangential velocity needed to keep the person from slipping downward? 1p gR 2 p 2. v = µ 2 g R 1. v =
fsmax
µ m v2 = µN = ≥ mg, R
the minimum speed required to keep the person supported is at the limit of this inequality, which is 2 µ m vmin = m g, or R µ ¶1 gR 2 vmin = . µ
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004 005 (part 2 of 2) 0 points Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimum necessary. Find the magnitude of the frictional force between the person and the wall. 1. F =
m v2 µR
3
Noting that the maximum centripetal force is required when the race car is moving at the maximum allowed speed without slipping, we have 2 m vmax = µs m g R p =⇒ vmax = µs g R q = (0.293)(9.8 m/s2 )(59.1 m) = 13.0269 m/s .
2. F = m g correct m v2 3. F = R µ m v2 4. F = R µ m v2 R m v2 = µmg + R 2 µmv − mg = R mg = µ m v2 = − µmg R
5. F = m g + 6. F 7. F 8. F 9. F
10. F = µ m g Explanation: The vertical friction force must equal m g, in order to balance the force of gravity and not have any acceleration in the vertical direction. 006 (part 1 of 2) 5 points A car of mass 629 kg travels around a flat, circular race track of radius 59.1 m. The coefficient of static friction between the wheels and the track is 0.293. The acceleration of gravity is 9.8 m/s2 . What is the maximum speed v that the car can go without flying off the track? Correct answer: 13.0269 m/s. Explanation: When the race car is going around a curved track, the maximum centripetal force that can be applied by the frictional force is fs = µs mg.
007 (part 2 of 2) 5 points The same car now travels on a straight track and goes over a hill with radius 194 m at the top. What is the maximum speed that the car can go over the hill without leaving the road? Correct answer: 43.6028 m/s. Explanation: m v2 = mg − N r where N is the normal force acting on the car from the ground. The car will fly off the ground just when N = 0 so the maximum speed allowed will be √ vmax = g r q = (9.8 m/s2 )(194 m) = 43.6028 m/s .
008 (part 1 of 4) 3 points The following figure shows a Ferris wheel that rotates 5 times each minute and has a diameter of 20 m. The acceleration of gravity is 9.8 m/s2 .
What is the centripetal acceleration of a rider? Correct answer: 2.74156 m/s2 .
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004 Explanation: The period of the Ferris wheel is T = 60 s/5 = 12 s . The speed of the wheel is 2πr T 2 π (10 m) = 12 s = 5.23599 m/s ,
v=
so the centripetal acceleration is a=
4
What force (magnitude) does the seat exert on a rider when the rider is halfway between top and bottom? Correct answer: 620.751 N. Explanation: In this case, the force exerted by the seat has two components: the vertical one balancing the gravity and the horizontal one providing the centripetal force. Thus we have p Fm = m g 2 + a2 q = (61 kg) (9.8 m/s2 )2 + (2.74156 m/s2 )2 = 620.751 N .
v2
r (5.23599 m/s)2 = 10 m = 2.74156 m/s2 .
009 (part 2 of 4) 3 points What force does the seat exert on a 61 kg rider at the lowest point of the ride? Correct answer: 765.035 N. Explanation: The force exerted by the seat balances the gravity and provides the centripetal force, so Fl = m [g + a] = (61 kg) (9.8 m/s2 + 2.74156 m/s2 ) = 765.035 N 010 (part 3 of 4) 2 points What force does the seat exert on a 61 kg rider at the highest point of the ride? Correct answer: 430.565 N. Explanation: The gravity is partly balanced by the force exerted by the seat and this resultant provides the centripetal force, so Fl = m [g − a] = (61 kg) (9.8 m/s2 − 2.74156 m/s2 ) = 430.565 N . 011 (part 4 of 4) 2 points
012 (part 1 of 1) 0 points A car traveling on a straight road at 7.9 m/s goes over a hump in the road. The hump may be regarded as an arc of a circle of radius 10 m. The acceleration of gravity is 9.8 m/s2 . What is the apparent weight of a(n) 381 N woman in the car as she rides over the hump? Correct answer: 138.365 N. Explanation: There are two forces acting on the woman of mass m. One is the gravity, W; the other is the supporting force from the chair, which is equal to the apparent weight Wa in magnitude. Their resultant force provides the centripetal force so W − W a = Fc =
m v2 r
then W v2 gr (381 N) (7.9 m/s)2 = 381 N − (9.8 m/s2 ) (10 m) = 138.365 N .
Wa = W − F c = W −
013 (part 1 of 2) 0 points A planet similar to the Earth has a radius 7.6 × 106 m and has an acceleration of gravity of 10 m/s2 on the planet’s surface. The planet rotates about its axis with a period of
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004 25 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, what is the new period? Correct answer: 1.52154 h. Explanation: When the apparent weight is zero, the gravity is equal to the centripetal force µ ¶2 2π 2 mg = mω R = m R Tf so this faster period, Tf , is s R Tf = 2 π g s 7.6 × 106 m = 2π 10 m/s2 = 5477.55 s = 1.52154 h .
Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass, as shown below. Each star has mass M and speed v.
v M v
Which of the following is a correct relationship among these quantities? 1. v 2 = 2. v 2 = 3. v 2 =
014 (part 2 of 2) 0 points By what factor would the speed of the object be increased when the planet is rotating at the higher speed? Correct answer: 16.4307 . Explanation: The velocity is
4. v 2 =
The period is T = 25 h. We know the ratio of vf the corresponding velocities, R = , is the v inverse of two different periods, which is, for this case, R=
vf T = v Tf 25 h 1.52154 h = 16.4307 .
=
015 (part 1 of 1) 0 points Given: G is the universal gravitational constant.
M
D
where R is the mean radius of the planet.
distance 2πR = . v= time T
5
5. v 2 = 6. v 2 = 7. v 2 =
GM D 4GM D 2GM D 2 G M2 D GM D2 4 G M2 D GM correct 2D
8. v 2 = M G D Explanation: From the basic formulae for the circular orbital movement, the centripetal acceleration v2 is a = . D 2 Using the Newton’s second law of motion, F we know the acceleration is a = , where F M is the force between two stars and is totally supplied by the universal force. So we obtain F GM 2 v2 =a= = D M D2
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004 =⇒
v2 =
GM . 2D
016 (part 1 of 2) 5 points On the way to the moon the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s. Determine the distance of this point from the center of the Earth. The masses of the Earth and the Moon are respectively 5.37 × 1024 kg and 7.36 × 1022 kg. The distance from the Earth to the Moon is 3.66 × 108 m. Correct answer: 3.27642 × 108 m. Explanation: If re is the distance from this point to the center of the Earth and rm is the distance from this point to the center of the Moon, then from the formula G m Mm G m Me = 2 2 re rm we obtain rm q= = re
r
Mm Me
s
7.36 × 1022 kg 5.37 × 1024 kg = 0.117072 . =
On the other hand, re + r m = R . Eliminating rm from the last two equalities, we obtain R q+1 3.66 × 108 m = 0.117072 + 1 = 3.27642 × 108 m .
re =
017 (part 2 of 2) 5 points What is the acceleration due to the Earth’s gravity at this point? The universal gravitational constant G is 6.672 × 10−11 N m2 /kg2 . Correct answer: 0.00333757 m/s2 . Explanation:
6
From the relation a=
F G Me = m re2
we obtain that the acceleration a due to the Earth’s gravity at this point is a = 6.672 × 10−11 N m2 /kg2 5.37 × 1024 kg × (3.27642 × 108 m)2 = 0.00333757 m/s2 . 018 (part 1 of 3) 0 points Given: A solar system similar to our Sun and Earth, where Mearth Rearth Msun sun Rearth
= 4.09 × 1024 kg = 6.38 × 106 m = 1.84 × 1030 kg = 1.56 × 1011 m .
Determine the magnitude of the change ∆F in gravitational force that the Sun exerts on a 73.1 kg woman standing on the equator at noon and midnight. Hint: Since ∆r is so small, use differentials. Assume: The Sun and the Earth are the only masses acting on the woman. Correct answer: 6.03305 × 10−5 N. Explanation: Basic Concepts The gravitational force exerted by m2 on m1 is ~ 21 = −G m1 m2 ˆr12 F (1) 2 r12 and the magnitude of the force is m1 m2 F =G . (2) r2 If we want to find a change ∆F due to a change ∆r, we write ∆F dF ∆r ≈ ∆r (3) ∆r dr since using “differentials” essentially amounts to approximating ∆F =
dF ∆F ≈ ∆r dr
(4)
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004 for small changes. Solution: Differentiating F with respect to r, we find dF m Ms = −2 G (5) dr r3 so, using differential approximations, m Ms ∆F ≈ −2 G ∆r . (6) r3 The change in distance ∆r is equal to the twice the radius of the Earth (i.e., the diameter of the Earth), since the woman moves from the closest point (noon) to the furthest point (midnight). ∆r = 2 RE = 2 (6.38 × 106 m) = 1.276 × 107 m . Therefore m Ms ∆r . (6) ∆F ≈ −2 G r3 ≈ −2 (6.67259 × 10−11 N m2 /kg2 ) (73.1 kg) (1.84 × 1030 kg) × (1.56 × 1011 m)3 × (1.276 × 107 m) = −6.03305 × 10−5 N |∆F | = 6.03305 × 10−5 N . 019 (part 2 of 3) 0 points Determine the fractional percent change in ∆F the Sun’s gravitational force % on the F woman due to the rotation of the Earth in the 12 hours between noon and midnight. Assume: The Sun and the Earth are the only masses acting on the woman. Correct answer: 0.016359 %. Explanation: Using Eqs. (2) and (6), the fractional change is m Ms −2 G ∆r ∆F r3 = (7) m Ms F G r2 2 ∆r =− r 2 (1.276 × 107 m) =− 1.56 × 1011 m = −0.00016359 ,
7
so the percentage decrease is ¯ ¯ ¯ ¯ ∆F ¯ ¯¯ ¯ ¯ = ¯−0.00016359¯¯ 100 ¯ F ¯ = 0.016359 % . 020 (part 3 of 3) 0 points This part is the same as the previous part except we are concerned with the Moon’s gravitation force instead of the Sun’s gravitational force on the woman. Given: A system similar to our Earth and Moon, where Mearth Rearth Mmoon moon Rearth
= 4.09 × 1024 kg = 6.38 × 106 m = 7.22 × 1022 kg = 3.89 × 108 m .
Determine the fractional percent change in ∆F % on the the Moon’s gravitational force F woman due to the rotation of the Earth in the 12 hours between noon and midnight. Assume: The Earth and the Moon are the only masses acting on the woman. Correct answer: 6.56041 %. Explanation: The fractional change is m Mmoon ∆r 3 ∆F r = m Mmoon F G r2 2 ∆r =− r 2 (1.276 × 107 m) =− 3.89 × 108 m = −0.0656041 , −2 G
so the percentage decrease is ¯ ¯ ¯ ¯ ∆F ¯ ¯¯ ¯ ¯ = ¯−0.0656041¯¯ 100 ¯ F ¯ = 6.56041 % .
