Question 1

A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 7.00 revolutions every 10.0 seconds. What is the magnitude of the acceleration of the ball? A) 67.9 m/s² B) 38.7 m/s² C) 29.3 m/s² D) 14.8 m/s² E) 74.2 m/s²

Accepted Answer

The acceleration of an object moving in a circle is given by the equation a = v^2/r, where v is the linear speed of the object and r is the radius of the circle. To find the linear speed, we can use the formula v = 2πr/T, where T is the period of the motion (the time it takes to make one complete revolution). In this case, T = 10/7 seconds (since the ball makes 7 revolutions in 10 seconds), so v = 2π(2)/10/7 = 8π/5 m/s. Plugging in v = 8π/5 m/s and r = 2.00 m, we get a = (8π/5)^2/2 = 38.7 m/s^2. Therefore, the correct answer is B.

Question 2

A wind farm generator uses a two-bladed propeller (see figure)mounted on a pylon at a height of 20 m.The length of each propeller blade is 12 m.A small piece from the tip of the propeller breaks off when the propeller is vertical.At that instant,the period of the motion of the propeller is 1.2 s.The fragment flies off horizontally,falls with negligible air resistance,and strikes the ground at P.
(a) How far is point P from the base of the pylon?
(b) At what angle with respect to the vertical is the fragment moving just as it strikes the ground at P?

Accepted Answer

(a)160 m 
(b)68°

Question 3

An object moves in a circle of radius R at constant speed with a period T.If you want to change only the period in order to cut the object's acceleration in half,the new period should be
A) T/4.
B) T/2.
C) T/ 
D) T 
E) 4T.

Accepted Answer

D

Question 4

An aircraft performs a maneuver called an "aileron roll." During this maneuver,the plane turns like a screw as it maintains a straight flight path,which sets the wings in circular motion.If it takes it 35 s to complete the circle and the wingspan of the plane is 11 m,what is the acceleration of the wing tip? A) 0.18 m/s² B) 0.99 m/s² C) 5.6 m/s² D) 1.0 m/s²

Accepted Answer

The answer of An aircraft performs a maneuver called an...

Question 5

A dumbbell-shaped object is composed by two equal masses,m,connected by a rod of negligible mass and length r.If I₁ is the moment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and I₂ is the moment of inertia with respect to an axis passing through one of the masses,it follows that A) I₁ = I₂. B) I₁ > I₂. C) I₂ > I₁.

Accepted Answer

The moment of inertia depends on the distribution of mass relative to the axis of rotation. For $I_{\text{center}}$, both masses contribute equally since the axis is equidistant from them. For $I_{\text{end}}$, one mass is at the axis (contributing 0 to the moment of inertia) and the other is twice as far from the axis compared to the $I_{\text{center}}$ case, leading to a larger moment of inertia due to the $r^2$ dependence in the moment of inertia formula ($I = mr^2$). Therefore, $I_{\text{end}} > I_{\text{center}}$.

Question 6

A satellite orbits the earth a distance of 1.50 × 10⁷ m above the planet's surface and takes 8.65 hours for each revolution about the earth.The earth's radius is 6.38 × 10⁶ m.The acceleration of this satellite is closest to A) 0.0690 m/s². B) 0.870 m/s². C) 1.91 m/s². D) 2.72 m/s². E) 9.80 m/s².

Accepted Answer

The answer of A satellite orbits the earth a distance...

Question 7

A horizontal disk rotates about a vertical axis through its center.Point P is midway between the center and the rim of the disk,and point Q is on the rim.If the disk turns with constant angular velocity,which of the following statements about it are true? (There may be more than one correct choice.)
A) P and Q have the same linear acceleration.
B) Q is moving twice as fast as P.
C) The linear acceleration of Q is twice as great as the linear acceleration of P.
D) The linear acceleration of P is twice as great as the linear acceleration of Q.
E) The angular velocity of Q is twice as great as the angular velocity of P.

Accepted Answer

,C

Question 8

When a rigid body rotates about a fixed axis,all the points in the body have the same
A) tangential speed.
B) angular acceleration.
C) tangential acceleration.
D) linear displacement.
E) centripetal acceleration.

Accepted Answer

When a rigid body rotates about a fixed axis, all the points in the body have the same angular acceleration. This is because the angular acceleration is determined solely by the torque applied to the body and the body's moment of inertia, which are constant for all points on the body. Therefore, choice B is the best answer. Choices A, C, D, and E are incorrect because they do not apply to all points in the body equally.

Question 9

A string is attached to the rear-view mirror of a car.A ball is hanging at the other end of the string.The car is driving around in a circle,at a constant speed.Which of the following lists gives all of the forces directly acting on the ball?
A) tension and gravity
B) tension
C) tension, gravity, and the centripetal force
D) tension, gravity, the centripetal force, and friction

Accepted Answer

The only forces directly acting on the ball are the tension in the string and gravity. The centripetal force is not a separate force, but rather the combination of all the other forces (in this case, tension and gravity) that keeps the ball moving in a circle. Friction is not present in this scenario unless specifically stated.

Question 10

A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the road is 0.600. What is the maximum speed at which the car can travel around this curve without sliding?
A) 29.6 m/s
B) 33.1 m/s
C) 24.8 m/s
D) 42.0 m/s
E) 37.9 m/s

Accepted Answer

The answer of A car enters a 300-m radius horizontal...

Question 11

A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center.How many rpm (rev/min)must this disk make so that the acceleration of all points on its rim is g/2?

Accepted Answer

The answer of A disk-shaped space station 175 m in...

Question 12

A 23 kg mass is connected to a nail on a frictionless table by a massless string 1.3 m long.There is no appreciable friction between the nail and the string.If the tension in the string is 51 N while the mass moves in a uniform circle on the table,how long does it take for the mass to make one complete revolution?
A) 4.8 s
B) 3.8 s
C) 4.5 s
D) 5.2 s

Accepted Answer

The answer of A 23 kg mass is connected to...

Question 13

Two particles,A and B,are in uniform circular motion about a common center.The acceleration of particle A is 8.5 times that of particle B.The period of particle B is 2.0 times the period of particle A.The ratio of the radius of the motion of particle A to that of particle B is closest to A) r_A/r_B = 2.1. B) r_A/r_B = 4.3. C) r_A/r_B = 18. D) r_A/r_B = 0.24. E) r_A/r_B = 17.

Accepted Answer

The acceleration in uniform circular motion is given by $a = \frac{v^2}{r} = \frac{4\pi^2r}{T^2}$, where $v$ is the velocity, $r$ is the radius of the motion, and $T$ is the period of the motion. Given that the acceleration of particle A is 8.5 times that of particle B and the period of particle B is 2 times the period of particle A, we can set up the following relationship using the formula for acceleration in terms of period and radius: $a_A = 8.5a_B$ and $T_B = 2T_A$. Substituting the acceleration formula, we get $\frac{4\pi^2r_A}{T_A^2} = 8.5\frac{4\pi^2r_B}{T_B^2}$. Since $T_B = 2T_A$, we can substitute $T_B$ with $2T_A$ in the equation, leading to the simplification and solving for the ratio $\frac{r_A}{r_B}$, which gives a value close to 2.1.

Question 14

A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be traveling when it tires just barely lose contact with the road when the car is at the top of the hill?
A) 45.5 m/s
B) 41.8 m/s
C) 34.3 m/s
D) 22.2 m/s
E) 27.6 m/s

Accepted Answer

For the tires to just barely lose contact with the road at the top of the hill, the centripetal force must equal the gravitational force acting on the car. The formula for centripetal force is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the car, $v$ is the velocity, and $r$ is the radius of curvature. The gravitational force is $F_g = mg$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). Setting these two forces equal gives $\frac{mv^2}{r} = mg$, which simplifies to $v^2 = rg$. Plugging in $r = 0.120 km = 120 m$ and $g = 9.8 m/s^2$, we find $v = \sqrt{120m \times 9.8 m/s^2} = \sqrt{1176} m/s \approx 34.3 m/s$.

Question 15

If you swing a bucket of water fast enough in a vertical circle,at the highest point the water does not spill out because an outward force balances the pull of gravity on the water.

Accepted Answer

At the highest point, the water does not spill out because the centripetal force required to keep it in circular motion is provided by the force of gravity acting on the water, not an outward force.

Question 16

An electrical motor spins at a constant 2857.0 rev/min. If the armature radius is 2.685 cm,what is the acceleration of the outer edge of the armature? A) 2403 m/s² B) 844.4 m/s² C) 241,100 m/s² D) 84.40 m/s²

Accepted Answer

The answer of An electrical motor spins at a constant...

Question 17

Point P in the figure indicates the position of an object traveling at constant speed clockwise around the circle.Which arrow best represents the direction of the acceleration of the object at point P? 
A)

B)

C) 
D)

E)

Accepted Answer

The answer of Point P in the figure indicates the...

Question 18

For an object in uniform circular motion,its velocity and acceleration vectors are always perpendicular to each other at every point in the path.

Accepted Answer

This is because acceleration in uniform circular motion is always directed towards the center of the circular path, while velocity is always tangent to the path. Perpendicular vectors have a dot product of zero, which means they are orthogonal to each other. Therefore, the velocity and acceleration vectors in uniform circular motion are always perpendicular to each other.

Question 19

If an object travels at a constant speed in a circular path,the acceleration of the object is
A) larger in magnitude the smaller the radius of the circle.
B) in the same direction as the velocity of the object.
C) smaller in magnitude the smaller the radius of the circle.
D) in the opposite direction of the velocity of the object.
E) zero.

Accepted Answer

The acceleration of an object moving in a circular path at constant speed is centripetal, directed towards the center of the circle. Its magnitude is inversely proportional to the radius of the circle, meaning it is larger for smaller radii.

Question 20

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?
A) 1.23
B) 0.816
C) 0.736
D) 0.952
E) 0.662

Accepted Answer

The centripetal force required to keep the car moving around the curve is provided by the force of static friction, which acts inward towards the center of the circle. The formula for the centripetal force is Fc = mv^2/r, where m is the mass of the car, v is the speed, and r is the radius of the curve. The force of static friction is given by fs ≤ μsFn, where μs is the coefficient of static friction, Fn is the normal force, and ≤ indicates that the static friction force cannot exceed a certain maximum value.
Setting these two equations equal, we get:
mv^2/r = μsFn
Since the car is not accelerating vertically, we know that the normal force is equal to the weight of the car, or Fn = mg, where g is the acceleration due to gravity. Substituting this into the above equation and solving for μs, we get:
μs = mv^2/(mgr) = v^2/(gr)
Plugging in the given values, we get:
μs = (40.0 m/s)^2 / (9.81 m/s^2 * 200 m) ≈ 0.816
Therefore, the minimum coefficient of static friction needed for the car to go around the curve without sliding is 0.816.

Quiz 11: Motion in a Circle

A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 7.00 revolutions every 10.0 seconds. What is the magnitude of the acceleration of the ball?

An object moves in a circle of radius R at constant speed with a period T.If you want to change only the period in order to cut the object's acceleration in half,the new period should be

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution about the earth.The earth's radius is 6.38 × 106 m.The acceleration of this satellite is closest to

When a rigid body rotates about a fixed axis,all the points in the body have the same

A string is attached to the rear-view mirror of a car.A ball is hanging at the other end of the string.The car is driving around in a circle,at a constant speed.Which of the following lists gives all of the forces directly acting on the ball?

A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the road is 0.600. What is the maximum speed at which the car can travel around this curve without sliding?

A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center.How many rpm (rev/min)must this disk make so that the acceleration of all points on its rim is g/2?

Two particles,A and B,are in uniform circular motion about a common center.The acceleration of particle A is 8.5 times that of particle B.The period of particle B is 2.0 times the period of particle A.The ratio of the radius of the motion of particle A to that of particle B is closest to

A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be traveling when it tires just barely lose contact with the road when the car is at the top of the hill?

If you swing a bucket of water fast enough in a vertical circle,at the highest point the water does not spill out because an outward force balances the pull of gravity on the water.

An electrical motor spins at a constant 2857.0 rev/min. If the armature radius is 2.685 cm,what is the acceleration of the outer edge of the armature?

Point P in the figure indicates the position of an object traveling at constant speed clockwise around the circle.Which arrow best represents the direction of the acceleration of the object at point P?

For an object in uniform circular motion,its velocity and acceleration vectors are always perpendicular to each other at every point in the path.

If an object travels at a constant speed in a circular path,the acceleration of the object is

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?

A satellite orbits the earth a distance of 1.50 × 10⁷ m above the planet's surface and takes 8.65 hours for each revolution about the earth.The earth's radius is 6.38 × 10⁶ m.The acceleration of this satellite is closest to