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.
θ
l
d
What is the speed of each seat? Correct answer: 7.81126 m/s. Explanation:
In the vertical direction we have T cosθ =m g ,
where T is the tension in the chain. In the horizontal direction we have
T sinθ = m v 2 r . Since r =`sinθ+ d 2 = (1.18 m) sin 43.7◦+11.4 m 2 = 6.51524 m, we have v=pg rtanθ = q (9.8 m/s2 ) (6.51524 m) tan 43.7◦ = 7.81126 m/s. 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 cosθ = (m+M)g T = (m+M)g cosθ = (10 kg + 62.5 kg) (9.8 m/s 2 ) cos 43.7◦ = 982.756 N. 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, andSP, are equal. ArcsQRandSP are semicircles.
S R
P 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 pointP?
1. tP O a tQ t tR tS tP 2. tP O a tQ t tR tS tP 3. tP O a tQ t tR tS tP
4. tP O a tQ t tR tS tP 5. tP O a tQ t tR tS tP correct Explanation:
DuringRSandP Q, the acceleration is zero because the velocity is a constant both in mag-nitude and in direction. During QRandSP, even though the magnitude of the velocity is a constant, it changes its direction, which will result in an acceleration of v
2
r , where vis the velocity and r is the radius of the arc. So the acceleration is fixed at a non-zero constant duringSP andQR.
004(part 1 of 2) 0 points
An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (see figure). The coefficient of static friction between the person and the wall is µand the radius of the cylinder is R.
ω R
What is the minimum tangential velocity needed to keep the person from slipping down-ward? 1.v = 1 2 p g R 2.v =µp2g R 3.v =p2µ g R 4.v = s g R µ correct 5.v = 1 µ p g R 6.v = 2pg R 7.v =µpg R 8.v =µp2π g R 9.v = 2µpg R 10.v =pg R Explanation: Basic Concepts: Centripetal force: F = m v 2 r
Frictional force: fs ≤µN =fsmax
Solution: The maximum frictional force due to friction isfmax = µN,whereN is the inward directed normal force of the wall of the cylinder on the person. To support the person vertically, this maximal friction force fsmax must be larger than the force of gravity m g so that the actual force, which is less than µN, can take on the value m g in the positive vertical direction. Now, the normal force supplies the centripetal acceleration v
2
R on the person, so from Newton’s second law,
N = m v 2 R . Since fsmax =µN = µ m v 2 R ≥m g ,
the minimum speed required to keep the per-son supported is at the limit of this inequality, which is µ m v2 min R =m g, or vmin = µ g R µ ¶12 .
005(part 2 of 2) 0 points
Suppose a person whose mass is m is being held up against the wall with a constant tan-gential velocity v greater than the minimum necessary.
Find the magnitude of the frictional force between the person and the wall.
1.F = m v 2 µ R 2.F =m g correct 3.F = m v 2 R 4.F = µ m v 2 R 5.F =m g+ µ m v 2 R 6.F =µ m g+ m v 2 R 7.F = µ m v 2 R −m g 8.F = m g µ 9.F = m v 2 R −µ m g 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 co-efficient 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 speedv 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 isfs =µsmg.
Noting that the maximum centripetal force is required when the race car is moving at the maximum allowed speed without slipping, we have m vmax2 R =µsm g =⇒ vmax =pµsg R = q (0.293)(9.8 m/s2)(59.1 m) = 13.0269 m/s. 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
r =m g− N
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 diame-ter of 20 m.
The acceleration of gravity is 9.8 m/s2.
What is the centripetal acceleration of a rider?
Explanation:
The period of the Ferris wheel is T = 60 s/5 = 12 s. The speed of the wheel is
v= 2π r T
= 2π(10 m) 12 s = 5.23599 m/s, so the centripetal acceleration is
a = v 2 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
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 balanc-ing the gravity and the horizontal one provid-ing the centripetal force. Thus we have Fm =mpg2+a2 = (61 kg) q (9.8 m/s2)2+ (2.74156 m/s2)2 = 620.751 N. 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 mag-nitude. Their resultant force provides the centripetal force so W − Wa =Fc = m v 2 r then Wa =W −Fc=W − W v2 g r = 381 N− (381 N) (7.9 m/s) 2 (9.8 m/s2) (10 m) = 138.365 N. 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 grav-ity of 10 m/s2 on the planet’s surface. The planet rotates about its axis with a period of
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 grav-ity is equal to the centripetal force
m g =m ω2R=m µ 2π Tf ¶2 R so this faster period, Tf, is
Tf = 2π s R g = 2π s 7.6×106 m 10 m/s2 = 5477.55 s = 1.52154 h.
whereRis the mean radius of the planet. 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 v= distance time = 2π R T .
The period isT = 25 h. We know the ratio of the corresponding velocities, R = vf
v , is the inverse of two different periods, which is, for this case, R= vf v = T Tf = 25 h 1.52154 h = 16.4307. 015(part 1 of 1) 0 points
Given: G is the universal gravitational con-stant.
Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass, as shown below. Each star has massM and speedv.
D v
v M M
Which of the following is a correct relation-ship among these quantities?
1.v2= G M D 2.v2= 4G M D 3.v2= 2G M D 4.v2= 2G M 2 D 5.v2= G M D2 6.v2= 4G M 2 D 7.v2= G M 2D correct 8.v2=M G D Explanation:
From the basic formulae for the circular or-bital movement, the centripetal acceleration isa = v
2
D 2
.
Using the Newton’s second law of motion, we know the acceleration is a = F
M, where F is the force between two stars and is totally supplied by the universal force.
So we obtain 2v2 D =a = F M = G M D2
=⇒ v2= G M 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× 1024kg and 7.36×1022kg. The distance from the Earth to the Moon is 3.66×108m. 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 Me r2 e = G m Mm r2 m we obtain q= rm re = r Mm Me = s 7.36×1022 kg 5.37×1024 kg = 0.117072. On the other hand,
re+rm =R .
Eliminating rm from the last two equalities, we obtain re = R q+ 1 = 3.66×10 8 m 0.117072 + 1 = 3.27642×108 m. 017(part 2 of 2) 5 points
What is the acceleration due to the Earth’s gravity at this point? The universal gravita-tional constantGis 6.672×10−11
N m2/kg2. Correct answer: 0.00333757 m/s2.
Explanation:
From the relation a = F
m = G Me
r2
e
we obtain that the acceleration a due to the Earth’s gravity at this point is
a= 6.672×10−11N m2/kg2 × 5.37×10 24 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 = 4.09×1024 kg Rearth = 6.38×10 6 m Msun = 1.84×1030 kg Rsunearth = 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 ∆ris 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 onm1is ~ F21=−Gm 1m2 r212 ˆ r12 (1)
and the magnitude of the force is F =Gm1m2
r2 . (2)
If we want to find a change ∆F due to a change ∆r, we write
∆F = ∆F ∆r ∆r≈
dF
dr ∆r (3) since using “differentials” essentially amounts to approximating
∆F ∆r ≈
d F
for small changes.
Solution: Differentiating F with respect to r, we find
dF
dr =−2G m Ms
r3 (5)
so, using differential approximations, ∆F ≈ −2Gm Ms
r3 ∆r . (6)
The change in distance ∆r is equal to the twice the radius of the Earth (i.e., the diam-eter of the Earth), since the woman moves from the closest point (noon) to the furthest point (midnight). ∆r= 2RE = 2 (6.38×106m) = 1.276×107m. Therefore ∆F ≈ −2Gm Ms r3 ∆r . (6) ≈ −2 (6.67259×10−11 N m2/kg2) × (73.1 kg) (1.84×10 30 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 the Sun’s gravitational force ∆F
F % on the 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 ∆F F = −2Gm Ms r3 ∆r Gm Ms r2 (7) =−2 ∆r r =−2 (1.276×10 7 m) 1.56×1011m =−0.00016359,
so the percentage decrease is ¯ ¯ ¯ ¯ ∆F F ¯ ¯ ¯ ¯ =¯¯ ¯−0.00016359 ¯ ¯ ¯100 = 0.016359 %. 020(part 3 of 3) 0 points
This part is the same as the previous part ex-cept we are concerned with the Moon’s gravi-tation force instead of the Sun’s gravigravi-tational force on the woman.
Given: A system similar to our Earth and Moon, where
Mearth = 4.09×1024kg Rearth = 6.38×106m Mmoon= 7.22×1022kg
Rearthmoon= 3.89×108m.
Determine the fractional percent change in the Moon’s gravitational force ∆F
F % on the 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 ∆F F = −2Gm Mmoon r3 ∆r Gm Mmoon r2 =−2 ∆r r =−2 (1.276×10 7 m) 3.89×108 m =−0.0656041, so the percentage decrease is
¯ ¯ ¯ ¯ ∆F F ¯ ¯ ¯ ¯ =¯¯ ¯−0.0656041 ¯ ¯ ¯100 = 6.56041 %.
