(c) what is the pails minimum speed at the top of the circle if no water is to spill out?
Ch 6, Circular Motion
F c = 1000 v 2 / r
Homework
Ch six, Round Motion
Ch6: 7, 9, 19, 21, 24, 48, 51, 63, 67
Questions 4, 5, 6, seven, 8, ix
| Hmwk, Ch 5 | Homework Assignment Page | PHY 1350's Home Folio | Hmwk, Ch 7 |
Additional problems from Serway's quaternary edition
(4 ed) 6.1 An motorcar moves at abiding speed over the crest of a colina. The commuter moves in a vertical circle of radius 18.0 m. At the tiptop of the hill, she notices that she barely remains in contact with the seat. Observe the speed of the vehicle.
(4 ed) 6.2 A auto rounds a banked curve equally in Figure 6.five. The radius of curvature of the road is R, the banking angle is
, and the coefficient of static friction is
.
(a) Determine the range of speeds the car can have without slipping upwards or down the road.
(b) Detect the minimum value formu such that the minimum speed is null.
(c) What is the range of speeds possible if R = 100 m,
Conceptual Questions
Q6.4 Why does mud fly off a rapidly turning auto tire?For mud to move in a circumvolve, at that place must be a (net) force on it -- directed toward the middle of the circle. The value of that force is Fc = m v2/ r. As the velocity (or speed) increases, that force must increase if the piece of mud is to keep to motility in a circle. As the velocity increases, the strength that holds the mud to the bicycle reaches its limit and the mud can no longer go around in a circle. When that limit is reached, the mud separates from the tire.
Q6.5 Imagine that you adhere a heavy object to i end of a spring and so whirl the leap and object in a horizontal circle (past belongings the costless finish of the spring). Does the leap stretch? If then, why? Hash out this in terms of the force causing the round motion.
For the heavy object to move in a circle, in that location must be a (net) force on it, directed toward the center. That strength is provided by the bound. For a bound to provide a forcefulness, information technology must exist stretched.
Q6.6 It has been suggested that rotating cylinders about 16 km in length and 8 km in bore be placed in space and used equally colonies. The purpose of the rotation is to simulate gravity for the inhabitants. Explicate this concept for producing an effective gravity.
To go along an inhabitant moving in a circle, the exterior border (or "floor") volition have to exert a force, directed toward the heart. This forcefulness is very much similar the strength that an ordinary floor exerts on a person and then this would give the illusion of gravity.
Q6.7 Why does a pilot tend to blackness out when pulling out of a steep dive?
Coming out of a steep swoop puts great peachy forces on everything -- including the airplane pilot's blood. These forces -- often chosen "inertial forces" -- cause the pilot's blood to movement from his brain to his lower body. If this happens plenty -- for long enough or if the forces are bully enough -- the encephalon is deprived of oxygen in the blood and the pilot blacks out.
This tends to happen when the forces are greater than about half-dozen times weight -- referred to equally "six g's". Pressure suits tin can "clasp" a pilot'southward legs and abdomane and go on the claret flowing to the encephalon. Some roller coasters go shut enough to these conditions and so that some victims -- er, I mean "guests" -- might black out. Some of these roller coasters pause for several seconds at the finish to let claret become dorsum to brains before letting the "guests" get off the coaster.
Q6.eight Describe a state of affairs in which a car commuter can take a centripetal acceleration just no tangential acceleration.
Uniform Circular Motion (UCM) is simply such a state of affairs. There will ever be a centripetal acceleration in circular movement considering of the changing direction. If the motion is uniform -- if the speed is constant -- in that location will be no tangential acceleration.
Q6.ix Describe the path of a moving object if its acceleration is abiding in magnitude at all times and
(a) perpendicular to the velocity; and
This is Uniform Round Motion (UCM).
(b) parallel to the velocity.
This is straight-line motility.
Issues from the current (fifth) edition of Serway and Beichner.
6.7 While ii astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to exist round and 100 km above the surface of the Moon. If the mass and radius of the Moon are 7.twoscore x 1022 kg and 1.70 x 10half-dozen m, make up one's mind
(a) the orbiting astronaut's acceleration,
(b) his orbital speed, and
(c) the menses of the orbit.
The orbital radius is
r = 100 km + 1.seventy 10 xhalf-dozen k = 0.x x 10half-dozen m + one.70 x 106 m = one.eighty ten xhalf dozen m
Fthousand = G 1000 m / rtwo = m ac
One thousand M / rtwo = a c
(6.672 x 10-11 N one thousandtwo / kg2 ) ( vii.40 x 1022 ) / ( 1.80 x 106 thousand)2 = a c
a c = 1.524 m / s2
a c = 52 / r
v2 = a c r = (1.524 m / s2) (i.80 x ten6 m)
v2 = 2.743 10 106 grand2 / s2
v = 1.656 x tenthree m / s
v = D / t = C / t = ii Î r / t
t = ii
r / v = 2
t = 6.829 x 103 due south [ min / 60 s] [ h / 60 min ]
t = i.897 h
6.9 A money placed 30 cm from the heart of a rotating horizontal turntable slips when its speed is 50 cm/s.
(a) What provides the forcefulness in the radial management when earlier the coin slips?
(b) What is the coefficient of static friction betwixt coin and turntable?
As always, don't start without good, clear diagrams. What are all the forces interim on the coin?
w is the weight, n is the "normal" force -- the strength perpendicular to the surface -- and Ffr is the strength of friction -- parallel to the surface. It is this friction force Ffr that provides the centeripetal forcefulness Fc.
Now, what is the coefficient of friction ?
Ffr = n
north = w = mg
Ffr = m grand
Ffr = Fc = m v2/r
chiliad 1000 = grand v2/r
chiliad = 52/r
= vtwo/(thousand r)
Be conscientious with the units. We know v in cm/s and we know r in cm. We commonly state thousand every bit 9.8 m/southward2. If you lot just plug in the associated numbers the answer will be horrendous -- because the units are not the same. We can make a big deal of converting units or nosotros can just utilise g as 980 cm/sii.
= (fifty cm/s)2/[(980 cm/due south2)(30 cm)]
= (fifty)2/[(980)(30)]
= 0.085
6.19 A pail of h2o is rotated in a vertical circle of radius one.0 m. What must be the minimum speed of the pail at the superlative of the circle if no water is to spill out?
At the tiptop of the circle, gravity pulls downwardly -- toward the centre of the circle -- with a force of w = grand g. If the water is only on the verge of spilling out, this the only force on the water so that must also be the centripetal forcefulness.
Fc = m v2/r = m g = due west
m v2/r = yard k
vii/r = m
v2 = g r
v2 = (nine.eight m/southii)(1.0 m)
v2 = 9.8 m2/southward2
five = 3.13 m/southward
vi.21 A roller-coaster vehicle has a mass of 500 kg when fully loaded with passengers.
(a) If the vehicle has a speed of 20.0 grand/south at point A, what is the force exerted by the track on the vehicle at this bespeak?
(b) What is the maximum speed the vehicle tin can accept at B and still remain on the rails?
(a) If the vehicle has a speed of xx.0 chiliad/southward at point A, what is the force exerted by the track on the vehicle at this point?
The forces on mass m, the roller-coaster vehicle, are its weight mg pointing downwards and the force the track exerts Fnorth pointing up. Therefore, the cyberspace force Fnet is
Fnet = Fn - m g
This cyberspace force is the centrepital force Fc,
Fc = grand v2 / r
Fnet = Fc
Fn - m g = k fiveii / r
Fnorth = (thousand v2 / r) + m g
Fn = thousand [( v2 / r) + g ]
Fn =(500 kg) [( (20 m/south)2 / x m) + 9.8 m/s2 ]
Fn =(500 kg) [ 49.8 chiliad/due south2 ]
Fnorth = 24,900 North
That is virtually five times its weight of 5,000 N.
(b) What is the maximum speed the vehicle tin have at B and even so remain on the track?
To "still remain on the track" means the normal strength Fnorthward has only gone to null, Fn = 0.
With the normal force equal to zero, there is just the weight mg available to supply the centripetal strength Fc,
Fc = Fnet
one thousand vtwo / r = m k
v2 / r = g
52 = g r = (9.8 chiliad / s2 ) (15 one thousand) = 147 mtwo / s2
v = 12.12 g /s
6.24 A 5.00-kg mass attached to a jump scale rests on a frictionless, horizontal surface as in Figure P6.26. The jump scale, attached to the front end of a boxcar, reads eighteen.0 Due north when the car is in motion.
(a) If the jump scale reads cipher when the car is at rest, determine the acceleration of the car while it is in motion.
(b) What will the bound scale read if the auto moves with abiding velocity?
(c) Describe the forces on the mass as observed by someone in the car and by someone at rest, outside the auto.
(a) If the spring calibration reads nothing when the automobile is at remainder, determine the acceleration of the car while it is in motion.
Fnet = Fcalibration = 18 Due north = (5 kg) a = ma
a = 18 N / five kg
a = 3.6 m / s2
(b) What will the spring scale read if the auto moves with constant velocity?
At constant velocity, the net force on the block must be zero. Merely the scale exerts a horizontal force on the block and so it must read zero.
(c) Draw the forces on the mass as observed by someone in the car and by someone at residual, outside the car.
An observer at rest, on the exterior sees the spring scale exerting a force to the correct which causes the block to accelerate to the correct. The force of the jump scale is the only horizontal strength on the block and it is the net force which causes the block to accelerate.
An observer riding along with the block inside the motorcar sees the block at rest in her reference frame. The spring scale exerts a force to the right just the block remains at residuum. Therefore, our ride-along observer will conclude that there is an additional forcefulness to the left interim on the block. We call this an "inertial strength" or a "ficticious force" due to the acceleration of the reference frame.
vi.48 In a hydrogen atom, the electron in orbit around the proton feels an attractive strength of about 8.20 x 10-viii N. If the radius of the "orbit" is 5.thirty x 10-11 m, what is the frequency in revolutions per second?
Fc = m five2 / r
five2 = Fc r / m = ( 8.twenty ten 10-8 N)(5.thirty ten 10-11 thousand) / (1.67 x 10-27 kg)
v2 = two.sixty x 10nine m2 / southward2
5 = 5.101 x ten4 m/s
five = r
f = 9.625 10 ten14 rad / sec [ rev / 2
f = 1.532 x 1014 rev / sec
6.51 An air puck of mass 0.25 kg is tied to a string and immune to revolve in a circle of radius 1.0 grand on a frictionless, horizontal table. The other stop of the string passes through a hole in the center of the table and a mass of 1.0 kg is tied to it (Fig P6.38). The suspended mass remains in equilibrium while the puck on the tabletop revolves. What are
(a) the tension in the string,
(b) the key strength exerted on the puck, and
(c) the speed of the puck?
What are (a) the tension in the string,
Look at the forces on the hanging mass. From those yous tin readily see that the tension in the string must equal the weight of the hanging mass,
T = k2 1000
T = (one.0 kg) ((nine.8 m/south2)
T = nine.8 Due north
(b) the central force exerted on the puck, and
The simply horizontal force acting on the puck is the tension in the string and so this tension is the central strength and is equal to the centripetal strength,
Fc = T = gii chiliad
Fc = T = nine.8 N
(c) the speed of the puck?
| Fc = grandi v2 / R = mtwo thousand | (0.25 kg) five2 / 1.0 m = 9.8 N |
| fiveii = (m2 / mone) g R | 5two = (nine.viii / 0.25) grandii / s2 = 39.2 mii / southtwo |
| v = SQRT [ (m2 / yard1) k R ] | 5 = half dozen.26 yard / s |
half-dozen.63 An amusement park ride consists of a large vertical cylinder that spins most its axis fast enough that any person within is held upwards against the wall when the floor drops away (Fig P6.49). The coefficient of static friction between the person and the wall is
south, and the radius of the cyliinder is R. 
6 Flags - St Louis - calls this ride "Tom's Twister". Midways at fairs frequently have 1 like this called "the Rotor". Another variation, often called "the Round-Upwardly" tilts victims -- I mean guests -- upwardly at an angle. Await for them.
(a) Prove that the maximum period of revolution necessary to keep the person from falling is T = (4 2 R s / chiliad)1/ii .
Fx = - Fn = - m v2 / R = - Fc
Fn = m 5ii / R
Ff = k g
Ff =
Ff =
v = C / T = ii
vtwo = iv
T2 =
T = [
(b) Obtain a numerical value fo T if R = 4.00 m and s = 0.iv.
T = [
T = [(0.4)(4
T = ii.5 s
(c) How many revolutions per minute does the cylinder make?
f = 1 / T = (1 rev) / (2.5 s) [ 60 due south / min ] = 24 rev / min
f = 24 rpm
Solutions to additional problems from Serway'southward fourth edition.
(iv ed) 6.ane An automobile moves at constant speed over the crest of a hill. The driver moves in a vertical circle of radius 18.0 m. At the top of the loma, she notices that she barely remains in contact with the seat. Discover the speed of the vehicle.
Since the driver "barely remains in contact with the seat", the normal forcefulness Fn, between hither and the seat, is zero. She moves in a circumvolve, and then the net forcefulness on her must provide the centripetal force and the merely strength available is her weight, mg.
Fc = m v2 / r
Fc = chiliad 1000
g 52 / r = m g
v2 / r = g
52 = m r = ( nine.8 m/s2 ) ( 18 m ) = 176.r m2 / s2
v = 13.28 thou / s
(4 ed) half dozen.2 A car rounds a banked bend as in Effigy half-dozen.v. The radius of curvature of the road is R, the cyberbanking angle is
(a) Determine the range of speeds the car can accept without slipping up or down the route.
(b) Notice the minimum value formu such that the minimum speed is nothing.
(c) What is the range of speeds possible if R = 100 m,
Get-go, a note of caution. It is very piece of cake -- nigh "automatic" -- to choose x- and y-axes like this,
with the 10-axis along the plane. But don't do that hither! The centripetal strength and acceleration will be directed toward the center of the circle. And the center of the circle does not lie along this x-centrality. Rather, choose the x-axis so the heart of the circle does lie on information technology.
Now, with these axes, the cyberspace force in the y-direction will exist zero, Fnet,y = 0 and the net strength in the x-direction will be the centripetal strength, Fcyberspace,ten = - Fc = - m vii / r
Friction always opposes the move so we must look at two possibilities for the friction force. We must apply separate gratuitous-body diagrams for the ii cases
when the car is near to slide "up" the depository financial institution
when the car is about to slide "down" the bank
Fnet,y = 0
Fn cos
Fnorth cos
Fn(cos
Fdue north = g g / (cos
Fcyberspace,x = - Fc = - m five2 / r
Finternet,10 = - Fn sin
thousand vtwo / r = Fn sin
m v2 / r = Fn sin
m v2 / r = Fn (sin
vii = Fn (sin
5ii = [one / (cos
5two = [one/(cos
v2 = [sin
Fnet,y = 0
Fn cos
Fnorthward cos
Fn(cos
Fn = k g / (cos
Fcyberspace,x = - Fc = - m vii / r
Fnet,x = - Fn sin
grand v2 / r = Fnorthward sin
chiliad v2 / r = Fnorthward sin
thousand v2 / r = Fn (sin
52 = Fdue north (sin
five2 = [1 / (cos
vtwo = [one/(cos
v2 = [sin
(c) Now, with specific numerical values, this is
vtwo = {[sin teno + 0.10 cos xo]/[cos 10o - 0.10 sin 10o]}(100 grand)(9.8 g/south2)
52 = {[0.174 + 0.x (0.985)]/[0.985 - 0.10 (0.174)]}(100 m)(ix.8 m/s2)
v2 = {[0.174 + 0.099]/[0.985 - 0.0174]}(100 m)(9.8 1000/southwardii)
vtwo = {[0.272]/[0.968]}(980 10002/s2)
v2 = 275 chiliadii/s2
v = 16.6 1000/s
This is the maximum speed.
(c) At present, with specific numerical values, this is
v2 = {[sin 10o - 0.10 cos 10o]/[cos teno + 0.10 sin 10o]}(100 1000)(ix.8 m/southward2)
v2 = {[0.174 - 0.10 (0.985)]/[0.985 + 0.10 (0.174)]}(100 one thousand)(ix.viii grand/s2)
52 = {[0.174 - 0.099]/[0.985 + 0.0174]}(100 chiliad)(ix.8 m/s2)
five2 = {[0.075]/[1.002]}(980 m2/s2)
v2 = 73.iii m2/due southii
v = 8.six m/s
This is the minimum speed.
(a) Determine the range of speeds the auto tin can have without slipping up or down the road.
(b) Find the minimum value for
Nosotros tin set five = 0. Actually, it is easier to set up v2 = 0
52 = [sin
sin
sin
(c) What is the range of speeds possible if R = 100 m,
These values are at the finish of the table above.
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(c) Doug Davis, 2001; all rights reserved
Source: https://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch06/Ch6.html
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