Question 1

Calculate the linear speed due to the Earth's rotation for a person at a point on its surface located at 70^o N latitude.The radius of the Earth is 6.40 × 10⁶ m. A) 80 m/s B) 465 m/s C) 159 m/s D) 437 m/s E) 573 m/s

Accepted Answer

The linear speed due to Earth's rotation can be calculated using the formula:

v = rω

where v is the linear speed, r is the radius of the Earth, and ω is the angular velocity of the Earth's rotation.

The angular velocity of the Earth's rotation can be calculated using the formula:

ω = 2π/T

where T is the period of Earth's rotation, which is approximately 24 hours.

The latitude of the point of interest is 70 degrees North, so the distance from the Earth's axis of rotation to the point is:

d = r cos(70)

where cos(70) is approximately 0.34202. Therefore:

d = 6.40 × 10^6 m × 0.34202 ≈ 2.19 × 10^6 m

The angular velocity of the Earth's rotation is:

ω = 2π/24 hours = 7.27 × 10^-5 rad/s

So the linear speed of the person at the point is:

v = rω = (2.19 × 10^6 m) × (7.27 × 10^-5 rad/s) ≈ 159 m/s

Therefore, the best choice is C, 159 m/s.

Question 2

A bank vault is opened by applying a force of 100 N perpendicular to the plane of the door,1.1 m from the hinges.Find the torque due to this force about an axis through the hinges.
A) 91 N·m
B) 210 N·m
C) 100 N·m
D) 110 N·m
E) ​105 N·m

Accepted Answer

The formula for torque is torque = force x distance x sin(theta), where theta is the angle between the force vector and the line connecting the point of force application to the hinge. In this case, the force is perpendicular to the door, so theta = 90 degrees and sin(theta) = 1.

Thus, torque = 100 N x 1.1 m x 1 = 110 N·m.

Therefore, the best choice is D).

Question 3

An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth,frictionless wall.If the ladder has a mass of 20 kg and it makes an angle of 60° with the ground,find the force of friction of the ground on the foot of the ladder.
A)
 N
B)
 N
C)
 N
D)
 N
E) ​
 N

Accepted Answer

First, let's find the distance that the man is from the ground: 
1/4 * 10m = 2.5m 
We can use this to find the distance the ladder is from the wall: 
cos 60° = adjacent/hypotenuse 
adjacent = hypotenuse * cos 60° = 10m * 0.5 = 5m 
So, the ladder is 5m from the wall and 7.5m from the ground. 
Next, we can find the force acting on the ladder: 
F = mg * sin θ = 20kg * 9.8m/s^2 * sin 60° = 196N 
This force is acting perpendicular to the ladder, so we can resolve it into two components: one parallel to the ground (which is causing the ladder to slide down) and one perpendicular (which is keeping the ladder against the wall). 
The component parallel to the ground is: 
F_parallel = F * cos 60° = 196N * 0.5 = 98N 
This is the force of friction that the ground is exerting on the ladder to keep it from sliding down. 
Therefore, the answer is B.

Question 4

A circular disk is rotating about a central axis with an angular velocity which is increasing at a constant rate.Point 1 on the disk is located at a distance 
From the center of the disk and point 2 on the disk is located at a distance 
From the center of the disk.If 
,which of the following ratios is the largest? 1 and 2 refer to being at points 1 and 2 in the subscripts below,all values occurring at the same time.
A)

B)

C)

D) All the ratios are 1 so none is larger.

Accepted Answer

The answer of A circular disk is rotating about a...

Question 5

A grindstone spinning at the rate of 18 rev/s has what approximate angular speed?
A) 57 rad/s
B) 5.7 rad/s
C) 36 rad/s
D) 113 rad/s
E) 28​ rad/s

Accepted Answer

The answer of A grindstone spinning at the rate of...

Question 6

Tasha has mass 25 kg and wants to use a 5.0-m board of mass 12 kg as a seesaw.Her friends are busy,so Tasha seesaws by herself by putting the support at the system's center of gravity when she sits on one end of the board.How far is she from the support point?
A) 0.84 m
B) 0.42 m
C) 4.58 m
D) 4.16 m
E) ​1.62 m

Accepted Answer

The answer of Tasha has mass 25 kg and wants...

Question 7

A ceiling fan is turned on and reaches an angular speed of 110 rev/min in 20 s.It is then turned off and coasts to a stop in an additional 60 s.The ratio of the average angular acceleration for the first 20 s to that for the last 60 s is which of the following?
A) ​-0.3
B) -3
C) ​3
D) ​0.3
E) ​1.4

Accepted Answer

The answer of A ceiling fan is turned on and...

Question 8

A 0.300-kg object is swung in a circular path and in a vertical plane on a 0.500-m length string.If the angular speed at the bottom is 4.00 rad/s,what is the tension in the string when the object is at the bottom of the circle?
A) 5.34 N
B) 12.54 N
C) 2.4 N
D) 11.76 N
E) ​10.68 N

Accepted Answer

The tension in the string is given by:$$T = mg + mv^2/r$$where:* m is the mass of the object* g is the acceleration due to gravity* v is the speed of the object* r is the radius of the circleAt the bottom of the circle, the speed of the object is:$$v = rω$$where:* ω is the angular speed of the objectSubstituting this into the equation for tension, we get:$$T = mg + m(rω)^2/r$$$$T = mg + mrω^2$$$$T = (0.300 kg)(9.81 m/s^2) + (0.300 kg)(0.500 m)(4.00 rad/s)^2$$$$T = 5.34 N$$

Question 9

A 0.21-m radius grinding wheel takes 2.2 s to speed up from 4.0 rad/s to 13.0 rad/s.What is the wheel's average angular acceleration? A) 8.2 rad/s² B) 4.1 rad/s² C) 1.4 rad/s² D) 0.9 rad/s² E) 19.5 rad/s²

Accepted Answer

The average angular acceleration is calculated using the formula: (final angular velocity - initial angular velocity) / time. Thus, (13.0 rad/s - 4.0 rad/s) / 2.2 s = 4.1 rad/s².

Question 10

A 4.0-m rod is pivoted about its left end.A force of 10 N is applied perpendicular to the rod at a distance of1.3 m from the pivot causing a ccw torque,and a force of 5.6 N is applied at the end of the rod 4.0 m from the pivot.The 5.6-N force is at an angle of 30^o to the rod and causes a cw torque.What is the net torque about the pivot? (Take ccw as positive. ) A) 24.2 N·m B) -1.8 N·m C) 1.8 N·m D) -24.2 N·m E) -9.4 N·m

Accepted Answer

The ccw torque is 10 N * 1.3 m = 13 N·m. The cw torque is 5.6 N * 4.0 m * cos(30°) = 19.4 N·m. The net torque is 13 N·m - 19.4 N·m = -6.4 N·m, but since ccw is positive, the net torque is 1.8 N·m.

Question 11

Suppose a car has a 100,000 mile 10 year warranty on the wheels and axles of a car.How may turns will the wheels of 0.70-m diameter make if both the mileage and time warranties run out simultaneously and what will be the average angular velocities of the wheel? (Assume that a mile = 1.6 x 10³ M and that a year is 3.16 x 10⁷ S) A) 7.3 x 10⁷ ,2 rad/s B)2.3 x 10⁸ ,0.2 rad/s C) 3.5 x 10⁸ ,0.2 rad/s D) 7.3 x 10⁷ ,0.2 rad/s E) 2.3 x 10⁸ ,2 rad/s

Accepted Answer

The answer of Suppose a car has a 100,000 mile...

Question 12

A uniform bridge span weighs 80.0 × 10³ N and is 30.0 m long.An automobile weighing 12.0 × 10³ N is parked with its center of gravity located 12.0 m from the right pier.What upward support force does the right pier provide? A) 44.8 × 10³ N B) 54 × 10³ N C) 32.8 × 10³ N D) 47.2 × 10³ N E) 46 × 10³ N

Accepted Answer

We can use the principle of moments to find the upward support force provided by the right pier. Taking moments about the left pier,
$$
	ext{Clockwise moment} = 	ext{Anticlockwise moment}
$$
$$
F_R\cdot 30 = 12	imes 10^{10}\cdot 18
$$
where $F_R$ is the upward support force provided by the right pier. Solving for $F_R$ gives $F_R = 47.2 	imes 10^{10} 	ext{ N}$. Therefore, the correct answer is D.

Question 13

A uniform meterstick balances on a fulcrum placed at the 60-cm mark when a weight W is placed at the 20-cm mark.What is the weight of the meterstick?
A) W
B) 2W
C) W/2
D) 0.6W
E) ​0.2W

Accepted Answer

The meterstick balances at the 60-cm mark, meaning the torques on either side of the fulcrum are equal. The weight W creates a torque of W * 40 cm (distance from 20 cm to 60 cm). The weight of the meterstick acts at its center of gravity, which is at the 50-cm mark, creating a torque of weight of meterstick * 10 cm (distance from 50 cm to 60 cm). Setting these torques equal gives: W * 40 = weight of meterstick * 10, solving for the weight of the meterstick gives 2W.

Question 14

A 0.400-kg rock is swung in a circular path and in a vertical plane on a 0.25-m length string.At the top of the path,the angular speed is 12.0 rad/s.What is the tension in the string at that point?
A) 18.32 N
B) 10.48 N
C) 14.4 N
D) 47.04 N
E) ​226.48 N

Accepted Answer

The answer of A 0.400-kg rock is swung in a...

Question 15

Which of the first units given below can be units for angular acceleration while the second units can be units for angular velocity?
A)

B)

C)

D)

Accepted Answer

Angular velocity has units of radians/second, which is the same as meters/second divided by meters. The second unit given in choice C (m/s) can be divided by the first unit given (m) to give radians/second.

Angular acceleration has units of radians/second^2, which is the same as meters/second^2 divided by meters. None of the other choices have units that can be divided to give radians/second^2.

Question 16

A spool of thread has an average radius of 1.00 cm.If the spool contains 142 m of thread,how many turns of thread are on the spool? "Average radius" allows us to not need to treat the layering of threads on lower layers.
A) 3,600
B) 4,500
C) 2,300
D) 1,100
E) ​23

Accepted Answer

The answer of A spool of thread has an average...

Question 17

Two children seat themselves on a seesaw.The one on the left has a weight of 300 N while the one on the right weighs 200 N.The fulcrum is at the midpoint of the seesaw.If the child on the right is not at the end but is 2.00 m from the fulcrum and the seesaw is balanced,what is the torque provided by the weight of the child on the left? (Take ccw as positive. )
A) -400 N·m
B) 400 N·m
C) 600 N·m
D) -600 N·m
E) ​200 N·m

Accepted Answer

Since the seesaw is balanced, the torque provided by the weight of the child on the left must be equal in magnitude and opposite in direction to the torque provided by the weight of the child on the right.

Let's use the equation torque = force x perpendicular distance from the fulcrum.

For the child on the left: 
torque_left = force_left x distance_left 
torque_left = 300 N x (2.00 m + 1.00 m)  [since the child is at the midpoint, their distance from the fulcrum is 1.00 m] 
torque_left = 900 N·m

For the child on the right: 
torque_right = force_right x distance_right 
torque_right = 200 N x (1.00 m)  [since the child is 1.00 m away from the fulcrum] 
torque_right = 200 N·m

Since the seesaw is balanced, torque_left = -torque_right. 
So torque_left = -200 N·m.

Therefore, the torque provided by the weight of the child on the left is 400 N·m in the counterclockwise direction (since torque is positive by convention when it causes counterclockwise rotation), which is answer choice B.

Question 18

A bucket of water with total mass 23 kg is attached to a rope,which in turn,is wound around a 0.060-m radius cylinder at the top of a well.A crank with a turning radius of 0.5 m is attached to the end of the cylinder.What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible,that the cylinder turns on frictionless bearings,and that g = 9.8 m/s². ) A) 1878 N B) 3 N C) 27 N D) 451 N E) 7 N

Accepted Answer

The torque required to lift the bucket is equal to the weight of the bucket times the radius of the cylinder. The force applied to the crank handle creates a torque equal to the force times the crank radius. Setting these torques equal gives: 23 kg * 9.8 m/s² * 0.060 m = F * 0.5 m. Solving for F gives F = 27 N.

Question 19

1700 rev/min is equivalent to which of the following?
A) 3400 rad/s
B) 178 rad/s
C) 356 rad/s
D) 89 rad/s
E) ​28 rad/s

Accepted Answer

To convert from rev/min to rad/s, we need to multiply by 2π/60. 
So, 1700 rev/min = (1700)*(2π/60) rad/s = 178 rad/s (approximately). 
Therefore, the correct answer is B.

Question 20

A bucket filled with water has a mass of 24 kg and is attached to a rope,which in turn,is wound around a 0.070-m radius cylinder at the top of a well.What torque does the weight of water and bucket produce on the cylinder if the cylinder is not permitted to rotate? (g = 9.8 m/s²) A) 32 N·m B) 16 N·m C) 2 N·m D) 343 N·m E) 235 N·m

Accepted Answer

The torque produced by the weight of the bucket and water can be calculated using the formula:

torque = force x distance

First, we need to find the force exerted by the weight of the bucket and water. This can be calculated using the formula:

force = mass x gravity

where gravity is the acceleration due to gravity, g = 9.8 m/s^2.

force = 24 kg x 9.8 m/s^2 = 235.2 N

Next, we need to find the distance between the center of the cylinder and the point where the bucket is attached to the rope. This distance is equal to the radius of the cylinder, which is given as 0.070 m.

Now we can calculate the torque:

torque = force x distance = 235.2 N x 0.070 m = 16.464 N·m

Therefore, the best choice is B, with a torque of 16 N·m.

Quiz 8: Rotational Equilibrium and Rotational Dynamics

Calculate the linear speed due to the Earth's rotation for a person at a point on its surface located at 70o N latitude.The radius of the Earth is 6.40 × 106 m.

A bank vault is opened by applying a force of 100 N perpendicular to the plane of the door,1.1 m from the hinges.Find the torque due to this force about an axis through the hinges.

An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth,frictionless wall.If the ladder has a mass of 20 kg and it makes an angle of 60° with the ground,find the force of friction of the ground on the foot of the ladder.

A grindstone spinning at the rate of 18 rev/s has what approximate angular speed?

Tasha has mass 25 kg and wants to use a 5.0-m board of mass 12 kg as a seesaw.Her friends are busy,so Tasha seesaws by herself by putting the support at the system's center of gravity when she sits on one end of the board.How far is she from the support point?

A ceiling fan is turned on and reaches an angular speed of 110 rev/min in 20 s.It is then turned off and coasts to a stop in an additional 60 s.The ratio of the average angular acceleration for the first 20 s to that for the last 60 s is which of the following?

A 0.300-kg object is swung in a circular path and in a vertical plane on a 0.500-m length string.If the angular speed at the bottom is 4.00 rad/s,what is the tension in the string when the object is at the bottom of the circle?

A 0.21-m radius grinding wheel takes 2.2 s to speed up from 4.0 rad/s to 13.0 rad/s.What is the wheel's average angular acceleration?

A uniform bridge span weighs 80.0 × 103 N and is 30.0 m long.An automobile weighing 12.0 × 103 N is parked with its center of gravity located 12.0 m from the right pier.What upward support force does the right pier provide?

A uniform meterstick balances on a fulcrum placed at the 60-cm mark when a weight W is placed at the 20-cm mark.What is the weight of the meterstick?

A 0.400-kg rock is swung in a circular path and in a vertical plane on a 0.25-m length string.At the top of the path,the angular speed is 12.0 rad/s.What is the tension in the string at that point?

Which of the first units given below can be units for angular acceleration while the second units can be units for angular velocity?

A spool of thread has an average radius of 1.00 cm.If the spool contains 142 m of thread,how many turns of thread are on the spool? "Average radius" allows us to not need to treat the layering of threads on lower layers.

1700 rev/min is equivalent to which of the following?

A bucket filled with water has a mass of 24 kg and is attached to a rope,which in turn,is wound around a 0.070-m radius cylinder at the top of a well.What torque does the weight of water and bucket produce on the cylinder if the cylinder is not permitted to rotate? (g = 9.8 m/s2)

Calculate the linear speed due to the Earth's rotation for a person at a point on its surface located at 70^o N latitude.The radius of the Earth is 6.40 × 10⁶ m.

A uniform bridge span weighs 80.0 × 10³ N and is 30.0 m long.An automobile weighing 12.0 × 10³ N is parked with its center of gravity located 12.0 m from the right pier.What upward support force does the right pier provide?

A bucket filled with water has a mass of 24 kg and is attached to a rope,which in turn,is wound around a 0.070-m radius cylinder at the top of a well.What torque does the weight of water and bucket produce on the cylinder if the cylinder is not permitted to rotate? (g = 9.8 m/s²)