Question 1

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 2

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 3

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

To find the acceleration of the wing tip, we can use the formula for centripetal acceleration:

a = v^2/r

where v is the velocity of the wing tip and r is the radius of the circle (half the wingspan).

From the information given, we know that the plane completes a full circle in 35 seconds. Since the flight path is straight, the distance traveled is equal to the circumference of the circle:

C = 2πr

Plugging in r = 11/2 = 5.5m and solving for v:

v = C/t = (2πr)/t = (2π*5.5)/35 = 0.99 m/s

Now we can find the acceleration:

a = v^2/r = (0.99^2)/5.5 = 0.18 m/s^2

Therefore, the answer is A.

Question 4

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 5

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 acceleration of the satellite is given by the formula:
a = v^2/r
where v is the velocity of the satellite and r is the distance between the satellite and the center of the earth.

To find the velocity of the satellite, we first need to find the period of the orbit:
T = 8.65 hours = 31140 seconds

Then we can use the formula for the circumference of a circle to find the distance traveled by the satellite in one orbit:
C = 2πr
distance = C = 2π(6.38 × 10^6 m + 1.50 × 10^7 m) = 1.12 × 10^8 m

Now we can find the velocity:
velocity = distance/time = 1.12 × 10^8 m/31140 s = 3608.7 m/s

Finally, we can plug in the values for v and r to find the acceleration:
a = v^2/r = (3608.7 m/s)^2/(6.38 × 10^6 m + 1.50 × 10^7 m) = 0.870 m/s^2

Therefore, the closest answer choice is B, 0.870 m/s^2.

Question 6

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 can be found using the equation a = v^2/r, where v is the velocity of the object and r is the radius of the circle.

To find the velocity of the ball, first calculate its angular velocity using the equation w = 2πn/t, where n is the number of revolutions and t is the time.

w = (2π)(7/10) = 4.4 rad/s

The linear velocity of the ball is then v = rw, where r is the radius of the circle.

v = (2.00 m)(4.4 rad/s) = 8.80 m/s

Finally, the acceleration of the ball is a = v^2/r.

a = (8.80 m/s)^2/2.00 m = 38.7 m/s^2

Therefore, the magnitude of the acceleration of the ball is 38.7 m/s^2, which corresponds to choice B.

Question 7

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 centripetal force on the mass is given by F = mv\^2/r, where m is the mass, v is the velocity, and r is the radius of the circle. In this case, the tension in the string provides the centripetal force:

F = T = mv\^2/r

Rearranging this equation gives:

v = sqrt(Tr/m)

The time it takes for the mass to make one complete revolution is the period T, which is given by:

T = 2πr/v

Substituting the expression for v gives:

T = 2πr/sqrt(Tr/m)

Plugging in the given values gives:

T = 2π(1.3 m)/sqrt(51 N * 1.3 m / 23 kg)

Solving for T gives:

T = 4.8 s

Therefore, the correct answer is A) 4.8 s.

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 two forces directly acting on the ball are tension (due to the string) and gravity. The centripetal force is not a separate force, but rather the net force acting towards the center of the circular path, which is the vector sum of tension and gravity. Friction is not directly acting on the ball, but rather on the tires of the car.

Question 10

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

To find the minimum coefficient of static friction, we need to use the centripetal force equation: $F_c = \frac{mv^2}{r}$. The force pushing the car outward is the force of friction, which is equal to $\mu_smg$, where $\mu_s$ is the coefficient of static friction, $m$ is the mass of the car, $g$ is the acceleration due to gravity, and $r$ is the radius of the curve. Setting these two equations equal to each other and solving for $\mu_s$, we get: $\mu_s = \frac{v^2}{rg}$. Plugging in the given values, we get: $\mu_s = \frac{(40.0	ext{ m/s})^2}{(200	ext{ m})(9.81	ext{ m/s}^2)} \approx \boxed{0.816}$.

Question 11

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 12

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

The answer of A wind farm generator uses a two-bladed...

Question 13

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

The answer of An object moves in a circle of...

Question 14

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 15

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 16

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 17

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 18

A 1.2-kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0.50 m/s.At the origin 0,the bomb explodes into two fragments.Fragment 1 has a mass of 0.40 kg and a speed of 0.90 m/s along the negative y-axis.In the figure,the energy released by the explosion is closest to 
A) 0.20 J.
B) 0.24 J.
C) 0.28 J.
D) 0.32 J.
E) 0.36 J.

Accepted Answer

The answer of A 1.2-kg spring-activated toy bomb slides on...

Question 19

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 20

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 true because the acceleration vector is directed towards the center of the circle while the velocity vector is tangential to the circle, forming a right angle between them.

Quiz 1: A: Foundations

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 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?

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 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?

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 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?

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 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 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?

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?

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

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?

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

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