Physics Chapter 5 Questions and Answers

why do airplanes bank when they turn? how would you compute the banking angle given its speed and radius of the turn?

airplanes bank when they turn because in order to turn, *there must be a force that will be exerted towards the center of a circle.* by tilting wings, *the lift force on the wings has a non-vertical component which points toward the center of the curve, providing the centripetal force.* the *banking angle can be computed from a free body diagram*. the sum of vertical forces must be 0 for the plane to execute a level turn, and so *F(lift)cosθ = mg*. *the horizontal component of the lifting force must provide the centripetal force to move the airplane in a circle* *F(lift)sinθ = m v^2/r —> (mg/cosθ)sinθ = m v^2/r —> tanθ = v^2/Rg*

will an object weigh more at the equator or at the poles? what two effects are at work? do they oppose each other?

an object *weighs more at the poles*, due to *2 effects which complement (not oppose) each other.* first of all, the Earth is slightly *flattened at the poles* and expanded at the equator, relative to a perfect sphere. thus the *mass at the poles is slightly closer to the center*, and so experiences a slightly larger gravitational force. secondly, *objects at the equator have a centripetal acceleration due to the rotation of the Earth* that objects at the poles DO NOT have. to provide that centripetal acceleration, *the apparent weight (the radially outward normal force of the Earth on an object) is slightly less than the gravitational pull inward.* so the *2 effects both make the weight of an object at the equator less than at the poles.*

the Earth moves faster in its orbit around the Sun in January than in July. is the Earth closer to the Sun in January or in July? explain. [note: this is not much of a factor in producing the seasons - the main factor is the tilt of the Earth's axis relative to the plane of its orbit]

by *Kepler's 2nd law*, the *Earth moves faster around the Sun when it is nearest the Sun.* Kepler's 2nd law says that *an imaginary line drawn from the Sun to the Earth* sweeps out equal areas in equal times*. so *when the Earth is close to the Sun, it must move faster to sweep out a given area than when the Earth is far from the Sun. thus the Earth is closer to the Sun in January.*

how many "accelerators" do you have in your car? there are at least 3 controls in the car which can be used to cause the car to accelerate. what are they? what acceleration do they produce?

the *3 major "accelerators"* are the *accelerator pedal, the brake pedal, and the steering wheel.* the *accelerator pedal*/gas pedal can be used to *increase speed* (by depressing the pedal) or to *decrease speed in combination with friction* (by releasing the pedal). the *brake pedal* can be used to *decrease speed* by depressing it. the *steering wheel* is used to change direction*, which also is an acceleration. there are some other controls which could also be considered accelerators. the parking brake can be used to decrease speed by depressing it. the gear shift lever can be used to decrease speed by downshifting. if the car has a manual transmission, then the clutch can be used to decrease speed by depressing it (friction will slow the car). finally, shutting the car off can be used to decrease its speed. *any changes in speed or direction means that an object is accelerating.*

when will your apparent weight be the greatest, as measured by a scale in a moving elevator: when the elevator accelerates downward, accelerates upward, is in freefall, moves upward at a constant speed? in which case would your weight be the least? when would it be the same as when you are on the ground?

the *apparent weight (the normal force) would be largest when the elevator is accelerating upward.* with a *positive acceleration, the normal force is greater than your weight.* the apparent weight would be the *least when in free fall, because there the apparent weight is 0, since a = -g.* when the elevator is moving *constant speed, your apparent weight would be the same as it is on the ground, since a = 0 and so FN = mg*

does an apple exert a gravitational force on the Earth? If so, how large a force? consider an apple (a) attached to a tree, and (b) falling.

the *apple DOES exert a gravitational force on the Earth*. by *Newton's 3rd law*, *the force on the Earth due to the apple is the same magnitude as the force on the apple due to the Earth* - the weight of the apple. *the force is also independent of the state of motion of the apple.* so for *both* a hanging apple and a falling apple, *the force on the Earth due to the apple is equal to the weight of the apple*

will the acceleration of a car be the same when the car travels around a sharp curve at a constant 60 km/h as when it travels around a gentle curve at the same speed? explain.

the *centripetal acceleration* for an object moving in circular motion is * inversely proportional to the radius of the curve*, given a constant speed *(a = v^2/r)*. so for a *gentle curve* (which means a large radius), *the acceleration is smaller*, while for a *sharp curve* (which means a small radius), the *acceleration is larger*.

a child on a sled comes flying over the crest of a small hill. his led does not leave the ground (he doesn't achieve "air"), but he feels the normal force between his chest and the sled decrease as he goes over the hill. explain this decrease using Newton's 2nd law.

when the child is on a level surface, the *normal force between his chest and the sled is equal to the child's weight*, and thus he has *no vertical acceleration*. when he goes over the hill, *the normal force on him will be reduced.* since the child is *moving on a curved path, there must be a net centripetal force towards the center of the path*, and so *the normal force does not completely support the weight.* write Newton's 2nd law for the radial direction, with inward as positive. *ΣFR = mg - FN = mv^2/r —> FN = mg - m v^2/r. * *we see that the normal force is reduced from mg by the centripetal force.*

explain how a runner experiences "free fall" or "apparent weightlessness" between steps.

when the runner has both feet off the ground, *the only force on the runner is gravity* - there is *no normal force from the ground* on the runner. this *lack of normal force is interpreted as "free fall" and apparent weightlessness"*

what keeps a satellite up in its orbit around the earth?

a satellite remains in orbit due to the combination of gravitational force on the satellite directed towards the center of the orbit and the tangential speed of the satellite. first, the proper tangential speed had to be established by some other force than the gravitational force. then if the satellite has the proper combination of speed and radius such that the force required for circular motion is equal to the force of gravity on the satellite, then the satellite will maintain circular motion.

a bucket of water can be whirled in a vertical circle without the water spilling out, even at the top of the circle when the bucket is upside down. explain.

for the *water to remain in the bucket*, there *must be a centripetal force forcing the water to move in a circle* along with the bucket. that centripetal force *gets larger with the tangential velocity of the water,* since *F(R) = mv^2/r*. *the centripetal force at the top of the motion comes from a combination of the downward force of gravity and the downward normal force* of the bucket on the water. if the bucket is *moving faster than some minimum speed, the water will stay in the bucket*. if the bucket is *moving too slow, there is insufficient force to keep the water moving in the circular path,* and it spills out.

would it require less speed to launch a satellite towards the east or toward the west? consider the Earth's rotation direction.

in order to orbit, *a satellite must reach an orbital speed relative to the center of the Earth.* since the *satellite is already moving eastward when launched (due to the rotation speed at the surface of the Earth), it requires less additional speed to launch it east to obtain the final orbital speed*

the mass of Pluto was not known until it was discovered to have a moon. explain how this discovery enabled an estimate of Pluto's mass.

let the mass of Pluto be M, the mass of the moon be m, the radius of the moon's orbit be R, and the period of the moon's orbit be T. then *Newton's 2nd law for the moon orbiting Pluto will be F = (GmM)/(R^2).* if that moon's orbit is a circle, then *the form of the force must be centripetal, and so F = mv^2/R.* equate these 2 expressions for the force on the moon, and *substitute the relationship for a circular orbit that v = 2πR/T.* *(GmM)/(R^2) = mv^2/R = 4π^2mR/T^2 --> M = 4π^2R^3/GT^2*. thus a value for the mass of Pluto can be calculated knowing the period and radius of the moon's orbit.

a girl is whirling a ball on a string around her head in a horizontal plane. she wants to let go at precisely the right time so that the hall will hit a target on the other side of the yard. when should she let go of the string?

she should let go of the string *when the ball is at a position where the tangent line to the circle at the ball's location, when extended, passes through the target's position*. that *tangent line indicates the direction of the velocity at that instant*, and if the centripetal force is *removed, the ball will follow that line horizontally.*

the sun's gravitational pull on the earth is much larger than the moon's. yet the moon's is mainly responsible for the tides. explain. (hint: consider the difference in gravitational pull from one side of the earth to the other.)

the *difference in force on the 2 sides of the Earth from the gravitational pull of either the Sun or the Moon is the primary cause of the tides.* that difference in force comes about from *the fact that the 2 sides of the Earth are a different distance away from the pulling body.* relative to the Sun, the *difference in distance (earth diameter) of the 2 sides from the Sun, relative to the average distance to the Sun, is given by 2Rearth/Rearth-to-sun = 8.5 x 10^-5*. the corresponding relationship between the *Earth and the Moon is 2Rearth/Rearth-to-moon = 3.3 x 10^-2*. since the *relative change in distance is much greater* for the *Earth-Moon combination*, we see that *the Moon is the primary cause of the Earth's tides.

the gravitational force on the moon due to the earth is only about half of the force on the moon due to the sun. why isn't the moon pulled away from the earth?

the Moon is not pulled away from the Earth because *both the Moon and the Earth are experiencing the same radial acceleration due to the Sun.* they *both have the same period around the Sun* because they are both, on average, *the same distance from the Sun*, and so *they travel around the Sun together.*

is the centripetal acceleration of Mars in its orbit around the Sun larger or smaller than the centripetal acceleration of the Earth?

the centripetal acceleration of Mars is *smaller than that of Earth*. the acceleration of each planet can be found by *dividing the gravitational force on each planet by the planet's mass.* the resulting acceleration is *inversely proportional to the square of the distance of the planet from the Sun*. since Mars is further from the Sun than the Earth is, *the acceleration of Mars will be smaller.* also see the F(on-planet) = G(MsunMplanet/R(sun-to-planet)^2) a(planet) = F(on-planet)/Mplanet = G(Mplanet/R(sun-to-planet)^2)

suppose a car moves at constant speed along a hilly road. where does the car exert the greatest and least forces on the road: (a) at the top of a hill, (b) at a dip between two hills, (c) on a level stretch near the bottom of a hill?

the force that the car exerts on the road is Newton's 3rd law reaction to the normal force of the road on the car, and so we can answer this question in terms of the normal force. *the car exerts the greatest force on the road at the dip between 2 hills*. there, *the normal force from the road has to both support the weight AND provide a centripetal upward force to make the car move in an upward curved path*. the car exerts the *least force* on the road *at the top of a hill*. we have all felt the "floating upward" sensation as we have driven over the crest of a hill. in that case, *there must be a net downward centripetal force to cause the circular motion, and so the normal force from the road does not completely support the weight*.

if the Earth's mass was double what it is, in what ways would the Moon's orbit be different?

the gravitational force on the Moon is given by *G(MearthMmoon/R^2)*, where R is the radius of the Moon's orbit. *this is a radial force, and so can be expressed as [(Mmoon)(vmoon^2)]/R.* this can be changed using the relationship vmoon = 2πR/T, where T is the orbital period of the Moon, to [(4π^2)(Mmoon)(R)]/T^2. *if we equate these 2 expressions for the force we get the following: G(MearthMmoon/R^2) = [(4π^2)(Mmoon)(R)]/T^2 --> R^3/T^2 = GMearth/(4π^2).* thus the mass of the earth determines the ratio R^3/T^2. *if the mass of the Earth were doubled, then the ratio R^3/T^2 would double, and so R'^3/T'^2 = R^3/T^2, where the primes indicated the "after doubling" conditions. for example, the radius might stay the same, and the period *decrease by a factor of 2^1/2, which means the speed increased by a factor of 2^1/2.* or the period might stay the same, and the *radius increase by a factor of 2^1/3, which means the speed increased by the same factor of 2^1/3.* or if both R and T were to double, keeping the speed constant, then R^3/T^2 would double. there are an infinite number of other combinations that would also satisfy the doubling of R^3/T^2.

which pulls harder gravitationally, the earth on the moon, or the moon on the earth? which accelerates more?

the gravitational pull is *the same in each case*, by Newton's 3rd law. *the magnitude of that pull is given by F = G(MearthMmoon/r(earth-moon)^2).* to find the acceleration of each body, *the gravitational pulling force is divided by the mass of the body.* since the *Moon has the smaller mass, it will have the larger acceleration*

why do bicycle riders lean inward when rounding a curve at high speed?

when a bicycle rider leans inward, *the bike tire pushes down on the ground at an angle.* the *road surface then pushes back* on the tire both vertically (*to provide the normal force which counteracts gravity*) and horizontally towards center of the curve (to provide the centripetal frictional force, enabling them to turn).

astronauts who spend long periods in outer space could be adversely affected by weightlessness. one way to simulate gravity is to shape the spaceships like a cylindrical shell that rotates, with the astronauts walking on the inside surface. explain how this simulate gravity. consider (a) how objects fall, (b) the force we feel on our feet, and (c) any other aspects of gravity you can think of.

the passengers are standing on the floor. (a) if a passenger held an object beside their waist and then released it, *the object would move in a straight line*, tangential to the circle in which the passenger's waist was moving when the object was released. In the figure, we see that *the released object would hit the rotating shell*, and so *fall to the floor*, but *behind the person*. the passenger might try to explain the motion by inventing some kind of "retarding" force on dropped objects, when *there really is no such force*. (b) *the floor exerts a centripetal force on the feet*, pushing them *toward the center*. this force has the same direction that a passenger would experience on earth, and so *it seems to the passenger that gravity must be pulling them "down"*. actually, the *passengers are pushing down on the floor, because the floor is pushing up from them*. (c) the "normal" way of playing catch, for example, *would have to change*. Since the artificial *gravity is not uniform*, passengers would *have to re-learn how to throw something across the room to each other*. there would *not be projectile motion* as we experience it on earth. also, if the cylinder were small, there might be a *noticeable difference in the acceleration* of our head vs. our feet. since the head is closer to the center of the circle then the feet, and both the head and the feet of the same period of rotation, *the centripetal acceleration (aR = 4π^2r/T^2) is smaller for the head.* this might *cause dizziness or a lightheaded feeling*.

sometimes people say that water is removed from clothes in a spin dryer by centrifugal force throwing the water outward. what is wrong with this statement?

the problem with the statement is that *there is nothing to cause an outward force*, and so the water is removed from the clothes is *NOT thrown outward*. Rather the *spinning drum pushes INWARD on the clothes and water.* but where there are *holes in the drum, the drum can't push on the water* and so the water is not pushed in. Instead the water moves tangentially to the rotation, out the holes, in a straight line, and so the water is separated from the clothes.

describe all the forces acting on a child riding a horse on a merry-go-round. which of these forces provides the centripetal acceleration of the child?

there are at *least 3 distinct major forces* on the child. the *force of gravity* is acting *downward* in the child. there is a *normal force* from the *seat of the horse acting upward* on the child. there *must be friction between the seat of the horse and the child or the child could not be accelerated by the horse*. it is that *friction that provides the centripetal acceleration*. there may be smaller forces as well, such as a reaction force on the child's hands of the child is holding on to part of the horse. *any force that has a radially inward component will contribute to the centripetal acceleration*.