Question 1

A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is half the radius for the outside. One car, car A, travels on the inside while another car of equal mass, car B, travels at equal speed on the outside of the curve. Which statement about resultant forces on the cars is correct?
A) The force on A is half the force on B.
B) The force on B is half the force on A.
C) The force on A is four times the force on B.
D) The force on B is four times the force on A.
E) There is no net resultant force on either as long as they stay on the road while turning.

Accepted Answer

The centripetal force required to keep a car moving in a circle is given by the formula $F = \frac{mv^2}{r}$, where $m$ is the mass of the car, $v$ is the velocity, and $r$ is the radius of the circle. Since both cars have the same mass and are traveling at the same speed, but car A is on a path with half the radius of car B, the force on car A is twice as much as the force on car B. This is because the force is inversely proportional to the radius.

Question 2

An iceboat is traveling in a circle on the ice. Halfway around the circle the sail and the steering mechanism fall off the boat. Which statement is correct?
A) The boat will continue traveling in the circle because there is no friction.
B) The boat will continue to travel in the circle because its velocity exerts a force on it.
C) The boat will move off on a line tangent to the circle because there is no force on it.
D) The boat will move off tangent to the circle because there is a force on it perpendicular to the boat directed to the outside of the circle.
E) The boat will move off to the outside perpendicular to the tangent line since a force directed to the outside of the circle always acts on the boat.

Accepted Answer

When the sail and steering mechanism fall off the boat, there is no longer any force acting on the boat to keep it traveling in a circle. The boat will therefore move off on a line tangent to the circle, since there is no force to change its direction. The absence of friction doesn't affect the boat's motion in this case because friction is not the only force acting on the boat. The boat's velocity does exert a force on it, but this force is in the direction of motion (tangential to the circle) and doesn't affect the boat's direction of travel. The other choices are also incorrect because they don't take into account the fact that there is no longer any force acting on the boat to keep it on a circular path.

Question 3

A boy on board a cruise ship drops a 30.0 gm marble into the ocean. If the resistive force proportionality constant is 0.500 kg/s, what is the terminal speed of the marble in m/s?
A) 0.147
B) 0.294
C) 0.588
D) 1.18
E) 2.35

Accepted Answer

The answer of A boy on board a cruise ship...

Question 4

A student is sitting on the right side of a school bus when it makes a right turn. We know that the force of gravity acts downwards and a normal force from the seat acts upwards. If the student stays in place when the bus turns, we also know that there must be
A) no other force on the student.
B) a force parallel to the seat directed forward on the student.
C) a force parallel to the seat directed to the left on the student.
D) a force parallel to the seat directed to the right on the student.
E) a force parallel to the seat in a direction between forward and left on the student.

Accepted Answer

When the bus turns right, the student experiences a change in direction, and according to Newton's first law, the student would continue moving in a straight line if there were no external forces acting on them. Therefore, to stay in place relative to the bus, there must be a force acting in the opposite direction of the turn. Since the turn is to the right, the force must be directed to the left. This force is provided by the frictional force between the student's body and the seat of the bus. This force is parallel to the seat and directed to the right, as indicated by choice D.

Question 5

Two small cylindrical plastic containers with flat bottoms are placed on a turntable that has a smooth flat surface. Canister A is empty; canister B contains lead shot. Each canister is the same distance r from the center. The coefficient of static friction between the canisters and the turntable is μ_s. When the speed of the turntable is gradually increased, A) only the lighter container slides outward off the turntable; the heavier one stays on. B) only the heavier container slides outward off the turntable; the lighter one stays on. C) both containers slide off the turntable at the same turntable speed. D) the lighter container slides inward. E) the heavier container slides inward.

Accepted Answer

The frictional force that keeps an object in circular motion is the same for both canisters since it depends on the coefficient of static friction and the normal force (which is the weight of the canister). The mass of the canisters affects the normal force, but since both are subjected to the same coefficient of static friction and are at the same distance from the center, they will start sliding at the same speed due to the frictional force being overcome at the same rate for both.

Question 6

A 4.0-kg mass attached to the end of a string swings in a vertical circle of radius 2.0 m. When the string makes an angle of 35° with the vertical as shown, the speed of the mass is 5.0 m/s. At this instant what is the magnitude of the force the string exerts on the mass? 
A) 50 N
B) 82 N
C) 89 N
D) 11 N
E) 61 N

Accepted Answer

The answer of A 4.0-kg mass attached to the end...

Question 7

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when it is at the highest point?
A)

B)

C)

D)

E)

Accepted Answer

The answer of A rock attached to a string swings...

Question 8

The equation below is the solution to a problem.  . The best physical representation of this equation is
A) a sphere of 2.00 kg mass under a 45.2 N tension when at the bottom of a vertical circle.
B) a sphere of 2.00 kg mass under a 45.2 N tension when at the side of a vertical circle.
C) a sphere of 2.00 kg mass under a 45.2 N tension when at the top of a vertical circle.
D) a sphere of 2.00 kg mass at any point on a horizontal circle.
E) a 2.00 kg gecko running on the ceiling with a speed of 8.00 m/s.

Accepted Answer

The answer of The equation below is the solution to...

Question 9

The equation below is the solution to a problem.  . The best physical representation of this equation is
A) a sphere of 2.00 kg mass under a 6.00 N tension when at the bottom of a vertical circle.
B) a sphere of 2.00 kg mass under a 6.00 N tension when at the side of a vertical circle.
C) a sphere of 2.00 kg mass under a 6.00 N tension when at the top of a vertical circle.
D) a sphere of 2.00 kg mass at any point on a horizontal circle.
E) a 2.00 kg gecko running on the ceiling with a speed of 8.00 m/s.

Accepted Answer

The given equation is the centripetal force equation: Fc = mv^2/r, where Fc represents the centripetal force, m represents the mass of the object moving in a circle, v represents its linear velocity, and r represents the radius of the circle.

Option A is incorrect because the sphere is at the bottom of a vertical circle, and at this point, the tension force will be greater than the weight of the object, so the net force will not be centripetal.

Option B is also incorrect because the sphere is at the side of a vertical circle, and there will be a component of its weight force acting tangentially, which again will result in net force not being centripetal.

Option C is correct because the sphere is at the top of a vertical circle and the tension force provided by the string will act in the downward direction, balancing the weight of the object, and providing the necessary centripetal force.

Option D is incorrect because the sphere is moving in a horizontal circle, and there is no vertical component of its motion.

Option E is unrelated to the given equation and is therefore incorrect.

Question 10

The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The car is on a level circular track of 50.0 m radius on a planet where  compared to Earth's  . The maximum safe speed on the track on the planet is ____ times as large as the maximum safe speed on Earth.
A) 0.250
B) 0.500
C) 1.00
D) 2.00
E) 4.00

Accepted Answer

The answer of The coefficient of static friction for the...

Question 11

For a plane to be able to fly clockwise in a horizontal circle as seen from above, in addition to exerting a force downwards on the air
A) it must be increasing its speed.
B) it must exert a force on the air that is directed to the plane's left side.
C) it must exert a force on the air that is directed to the plane's right side.
D) it does not need to exert a force: it must only move the wing flaps out.
E) it only needs to deflect the air without exerting any additional force on the air.

Accepted Answer

In order to fly in a horizontal circle, the plane needs to turn towards its left side. This requires the plane to create a centripetal force towards the center of the circle. Since the plane is already exerting a force downwards on the air, it needs to exert an additional force towards its left side which will create the necessary centripetal force. This can be achieved by deflecting the air using the ailerons on the wings to create a lift force on the left wing and a corresponding drag force on the right wing, which will turn the plane towards the left. Therefore, option B is the best choice.

Question 12

A race car traveling at 100 m/s enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tires and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct?
A) The race car will crash into the outer wall.
B) The race car will crash into the inner wall.
C) The car will stay in the center of the track.
D) The car will stay in the center of the track if the driver speeds up.
E) The car would stay in the center of the track if the radius were reduced to 200 m.

Accepted Answer

The centrifugal force acting on the car is given by Fc = mv^2/r, where m is the mass of the car, v is its velocity, and r is the radius of the turn. The maximum static frictional force that the track can exert on the car is given by Ff = μN, where μ is the coefficient of friction and N is the normal force on the car. In an unbanked turn, the normal force is simply the weight of the car (N = mg). Thus, the maximum speed that the car can safely travel around the turn is given by √(μgr), where g is the acceleration due to gravity.

Plugging in the given values, we get √(1.1 x 9.8 x 400) = 66.6 m/s. Since the car is traveling faster than this speed, it will not be able to make the turn and will crash into the outer wall.

Question 13

The equation below is the solution to a problem.  . The best physical representation of this equation is
A) a sphere of 2.00 kg mass under a 25.6 N tension when at the bottom of a vertical circle.
B) a sphere of 2.00 kg mass under a 25.6 N tension when at the side of a vertical circle.
C) a sphere of 2.00 kg mass under a 25.6 N tension when at the top of a vertical circle.
D) a sphere of 2.00 kg mass at any point on a horizontal circle.
E) a 2.00 kg gecko running on the ceiling with a speed of 8.00 m/s.

Accepted Answer

The answer of The equation below is the solution to...

Question 14

Frank says that if you release the string when swinging a ball in a horizontal circle, the ball flies out in the radial direction defined by the string at the instant you release the ball. John says that it flies out along a tangent line perpendicular to the string, and that it then drops straight down to the ground. Which one, if either, is correct?
A) Frank, because the centrifugal force is no longer counteracted by the string.
B) Frank, because balls naturally fly straight out.
C) John, because there is no centrifugal force.
D) John, because balls fall straight down when released.
E) Neither, because although there is no centrifugal force, and the ball's velocity is tangent to the circle at the instant of release, the ball then follows a parabolic trajectory.

Accepted Answer

The ball moves in a straight line tangent to the circle at the point of release due to inertia, and then follows a parabolic trajectory due to gravity.

Question 15

When a car goes around a circular curve on a level road without slipping,
A) no frictional force is needed because the car simply follows the road.
B) the frictional force of the road on the car increases when the car's speed decreases.
C) the frictional force of the road on the car increases when the car's speed increases.
D) the frictional force of the road on the car increases when the car moves to the outside of the curve.
E) there is no net frictional force because the road and the car exert equal and opposite forces on each other.

Accepted Answer

When a car goes around a circular curve on a level road without slipping, the frictional force of the road on the car increases when the car's speed increases. This is because the centripetal force required to keep the car moving in a circle increases with speed, and friction provides the necessary force for the car to maintain its circular path. The frictional force is proportional to the normal force between the tires and the road, which is the weight of the car in this case. Therefore, the frictional force increases with the weight of the car and the speed at which it is traveling.

Question 16

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when the string is in one possible horizontal position?
A)

B)

C)

D)

E)

Accepted Answer

The answer of A rock attached to a string swings...

Question 17

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when it is at the lowest point?
A)

B)

C)

D)

E)

Accepted Answer

At the lowest point, the tension in the string is providing the centripetal force necessary to keep the rock moving in a circle. Gravity is also acting on the rock, but it is not providing the centripetal force at this point in the motion. Therefore, the free body diagram should show tension pointing towards the center of the circle and gravity pointing downwards. Choice B correctly shows these two forces.

Question 18

A hornet circles around a pop can at increasing speed while flying in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are
A) straight down.
B) down and inwards.
C) down and outwards.
D) down and backwards.
E) down, backwards and outwards.

Accepted Answer

The answer of A hornet circles around a pop can...

Question 19

A hornet circles around a pop can at constant speed once per second in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are
A) straight down.
B) down and inwards.
C) down and outwards.
D) down and backwards.
E) down, inwards and backwards.

Accepted Answer

To maintain the circle at constant speed, the force applied to the air by the hornet's wings must create a centripetal force. Since the path of the hornet is curved inward towards the can, the force components must be directed outwards to create the necessary centripetal force. The only option that describes this is choice C, down and outwards.

Question 20

The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The car is on a level circular track of 50.0 m radius on a planet where  compared to Earth's  . If the car is to be able to travel at the same speed on the planet as on Earth, the radius of the track on the planet must be ____ times as large as the radius of the track on Earth.
A) 0.250
B) 0.500
C) 1.00
D) 2.00
E) 4.00

Accepted Answer

The maximum speed of the car on the track on Earth is determined by the maximum centripetal force it can sustain without slipping:

$F_{c,max} = \mu_smg$

where $\mu_s$ is the coefficient of static friction, $m$ is the mass of the car, $g$ is the acceleration due to gravity. At maximum speed $v$, the centripetal force acting on the car is provided by the frictional force:

$F_{c} = \mu_kmg$

where $\mu_k$ is the coefficient of kinetic friction. Equating the two forces and solving for $v$, we get:

$v = \sqrt{\mu_sgr}$

where $r$ is the radius of the track.

For the same value of $v$, the centripetal acceleration on the planet is 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11 times smaller than on Earth, because $g_{	ext{planet}} = 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11 g_{	ext{Earth}}$. Therefore, the radius of the track on the planet must also be 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11 times smaller in order to keep the centripetal force the same:

$r_{	ext{planet}} = \frac{1}{11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11} r_{	ext{Earth}}$

Substituting this expression into the previous equation, we get:

$v_{	ext{planet}} = \sqrt{\mu_s g_{	ext{planet}} r_{	ext{planet}}} = \sqrt{\mu_s (g_{	ext{Earth}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11) (r_{	ext{Earth}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11)} = \sqrt{\mu_s g_{	ext{Earth}} r_{	ext{Earth}}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11$

To keep $v_{	ext{planet}} = v_{	ext{Earth}}$, we need:

$\sqrt{\mu_s g_{	ext{Earth}} r_{	ext{Earth}}} = \sqrt{\mu_s g_{	ext{Earth}} r_{	ext{planet}}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11$

Squaring both sides and solving for $r_{	ext{planet}}$, we get:

$r_{	ext{planet}} = \frac{1}{11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11^2} r_{	ext{Earth}}$

Therefore, the radius of the track on the planet must be 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11^2 times smaller than on Earth, or 4 times smaller approximately. The answer is (E) 4.00.

Quiz 6: Circular Motion and Other Applications of Newtons Laws

An iceboat is traveling in a circle on the ice. Halfway around the circle the sail and the steering mechanism fall off the boat. Which statement is correct?

A boy on board a cruise ship drops a 30.0 gm marble into the ocean. If the resistive force proportionality constant is 0.500 kg/s, what is the terminal speed of the marble in m/s?

A student is sitting on the right side of a school bus when it makes a right turn. We know that the force of gravity acts downwards and a normal force from the seat acts upwards. If the student stays in place when the bus turns, we also know that there must be

A 4.0-kg mass attached to the end of a string swings in a vertical circle of radius 2.0 m. When the string makes an angle of 35° with the vertical as shown, the speed of the mass is 5.0 m/s. At this instant what is the magnitude of the force the string exerts on the mass?

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when it is at the highest point?

The equation below is the solution to a problem. . The best physical representation of this equation is

The equation below is the solution to a problem. . The best physical representation of this equation is

For a plane to be able to fly clockwise in a horizontal circle as seen from above, in addition to exerting a force downwards on the air

A race car traveling at 100 m/s enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tires and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct?

The equation below is the solution to a problem. . The best physical representation of this equation is

When a car goes around a circular curve on a level road without slipping,

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when the string is in one possible horizontal position?

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when it is at the lowest point?

A hornet circles around a pop can at increasing speed while flying in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are

A hornet circles around a pop can at constant speed once per second in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are