A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is placedthrough the right hole, which corresponds to the center of mass of the board. The board ¡s then held in place. 1. Predict the motion of the board after it is released from rest, Explain. 2. Check your prediction by observing the demonstration. a. Describe the angular acceleration of the board. Explain how you can tell. What does your answer imply about the ne, torque about the pivot? Explain. b. Describe the acceleration of the center of mass of the board. Explain how you can tell. What does your answer imply about the net force acting on the board? Explain. 3. Explain how your answers about net torque and net force in question 2 would change, if at all, if there is appreciable friction between the board and the pivot and the board remains at rest.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 4.3, Problem 1aT

A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is placedthrough the right hole, which corresponds to the center of mass of the board. The board ¡s then held in place.

1. Predict the motion of the board after it is released from rest, Explain.

Chapter 4.3, Problem 1aT, A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is

2. Check your prediction by observing the demonstration.

a. Describe the angular acceleration of the board. Explain how you can tell.

What does your answer imply about the ne, torque about the pivot? Explain.

b. Describe the acceleration of the center of mass of the board. Explain how you can tell.

What does your answer imply about the net force acting on the board? Explain.

3. Explain how your answers about net torque and net force in question 2 would change, if at all, if there is appreciable friction between the board and the pivot and the board remains at rest.

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

(1)

Expert Solution

To determine

To Predict: The motion of the board.

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 1

Formula used:

τ=r×F=r×mg

Calculation:

τ=r×F⇒τ=r×mg⇒τ=rmgsinθ (θ=90°)⇒τ=rmg

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

(2)

Expert Solution

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Explanation of Solution

Given:

τ=rmg

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 2

Formula used:

τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

Calculation:

From part 1, the net torque is τ=rmg

From the rotational equation of motion:

τ=Iα

α=dωdt

On comparing both the equations:

τ=rmg=Iα

α=rmgI

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is τ=Iα and the net force is mg .

Conclusion:

Hence, the angular acceleration is α=rmgI . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

(3)

Expert Solution

To determine

The effect of friction on the torque and force.

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

τ=r×F−frictional force

Fnet=mg−frictional force

Calculation:

τ=r×F=r×mgτ=rmgsinθ−μFfτ=rmg−μFf

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

04:28

Students have asked these similar questions

Please help

15 cm Two lenses lie 70 cm apart. The focal lengths are f₁ = 20 cm for lens 1 (converging), and f₂ = -51 cm for lens 2 (diverging). A 15 cm tall object sits 45 cm in front of lens 1. Note: the 50 and 100 cm markings on the line are not the positions of the lenses! a) First, draw the positions of the two lenses. Then draw a ray diagram with all the primary rays. Include the location of image 1 formed by lens 1 and the final image formed by lens 2. (4 points) 0 cm 50 cm 100 cm 150 cm b) For each of image 1 and the final image: are they real or virtual, upright or inverted (relative to the original object)? (1 point) c) Calculate the distance of the final image from the object. (2 points) d) Find the magnification and size in cm of the final image. (2 points) e) Do your calculations in parts c) and d) match your ray diagrams in part a)? (1 point)

A simple series circuit consists of a 150 Ω resistor, a 27.0 V battery, a switch, and a 2.00 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t =0s . Part A Part complete Part B Part complete Part C Find the electric flux at t =0.50ns. Express your answer in volt-meters. View Available Hint(s)for Part C Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type nothing V⋅m Part D Find the displacement current at t =0.50ns.

Answer 2

Textbook Question

Chapter 4.3, Problem 1aT

A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is placedthrough the right hole, which corresponds to the center of mass of the board. The board ¡s then held in place.

1. Predict the motion of the board after it is released from rest, Explain.

Chapter 4.3, Problem 1aT, A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is

2. Check your prediction by observing the demonstration.

a. Describe the angular acceleration of the board. Explain how you can tell.

What does your answer imply about the ne, torque about the pivot? Explain.

b. Describe the acceleration of the center of mass of the board. Explain how you can tell.

What does your answer imply about the net force acting on the board? Explain.

3. Explain how your answers about net torque and net force in question 2 would change, if at all, if there is appreciable friction between the board and the pivot and the board remains at rest.

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

(1)

Expert Solution

To determine

To Predict: The motion of the board.

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 1

Formula used:

τ=r×F=r×mg

Calculation:

τ=r×F⇒τ=r×mg⇒τ=rmgsinθ (θ=90°)⇒τ=rmg

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

(2)

Expert Solution

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Explanation of Solution

Given:

τ=rmg

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 2

Formula used:

τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

Calculation:

From part 1, the net torque is τ=rmg

From the rotational equation of motion:

τ=Iα

α=dωdt

On comparing both the equations:

τ=rmg=Iα

α=rmgI

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is τ=Iα and the net force is mg .

Conclusion:

Hence, the angular acceleration is α=rmgI . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

(3)

Expert Solution

To determine

The effect of friction on the torque and force.

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

τ=r×F−frictional force

Fnet=mg−frictional force

Calculation:

τ=r×F=r×mgτ=rmgsinθ−μFfτ=rmg−μFf

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

04:28

Answer 3

Textbook Question

Chapter 4.3, Problem 1aT

A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is placedthrough the right hole, which corresponds to the center of mass of the board. The board ¡s then held in place.

1. Predict the motion of the board after it is released from rest, Explain.

Chapter 4.3, Problem 1aT, A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is

2. Check your prediction by observing the demonstration.

a. Describe the angular acceleration of the board. Explain how you can tell.

What does your answer imply about the ne, torque about the pivot? Explain.

b. Describe the acceleration of the center of mass of the board. Explain how you can tell.

What does your answer imply about the net force acting on the board? Explain.

3. Explain how your answers about net torque and net force in question 2 would change, if at all, if there is appreciable friction between the board and the pivot and the board remains at rest.

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

(1)

Expert Solution

To determine

To Predict: The motion of the board.

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 1

Formula used:

τ=r×F=r×mg

Calculation:

τ=r×F⇒τ=r×mg⇒τ=rmgsinθ (θ=90°)⇒τ=rmg

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

(2)

Expert Solution

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Explanation of Solution

Given:

τ=rmg

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 2

Formula used:

τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

Calculation:

From part 1, the net torque is τ=rmg

From the rotational equation of motion:

τ=Iα

α=dωdt

On comparing both the equations:

τ=rmg=Iα

α=rmgI

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is τ=Iα and the net force is mg .

Conclusion:

Hence, the angular acceleration is α=rmgI . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

(3)

Expert Solution

To determine

The effect of friction on the torque and force.

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

τ=r×F−frictional force

Fnet=mg−frictional force

Calculation:

τ=r×F=r×mgτ=rmgsinθ−μFfτ=rmg−μFf

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

04:28

Answer 4

Textbook Question

Chapter 4.3, Problem 1aT

A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is placedthrough the right hole, which corresponds to the center of mass of the board. The board ¡s then held in place.

1. Predict the motion of the board after it is released from rest, Explain.

Chapter 4.3, Problem 1aT, A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is

2. Check your prediction by observing the demonstration.

a. Describe the angular acceleration of the board. Explain how you can tell.

What does your answer imply about the ne, torque about the pivot? Explain.

b. Describe the acceleration of the center of mass of the board. Explain how you can tell.

What does your answer imply about the net force acting on the board? Explain.

3. Explain how your answers about net torque and net force in question 2 would change, if at all, if there is appreciable friction between the board and the pivot and the board remains at rest.

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

(1)

Expert Solution

To determine

To Predict: The motion of the board.

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 1

Formula used:

τ=r×F=r×mg

Calculation:

τ=r×F⇒τ=r×mg⇒τ=rmgsinθ (θ=90°)⇒τ=rmg

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

(2)

Expert Solution

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Explanation of Solution

Given:

τ=rmg

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 2

Formula used:

τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

Calculation:

From part 1, the net torque is τ=rmg

From the rotational equation of motion:

τ=Iα

α=dωdt

On comparing both the equations:

τ=rmg=Iα

α=rmgI

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is τ=Iα and the net force is mg .

Conclusion:

Hence, the angular acceleration is α=rmgI . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

(3)

Expert Solution

To determine

The effect of friction on the torque and force.

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

τ=r×F−frictional force

Fnet=mg−frictional force

Calculation:

τ=r×F=r×mgτ=rmgsinθ−μFfτ=rmg−μFf

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

04:28

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(1)

Expert Solution

To determine

To Predict: The motion of the board.

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 1

Formula used:

τ=r×F=r×mg

Calculation:

τ=r×F⇒τ=r×mg⇒τ=rmgsinθ (θ=90°)⇒τ=rmg

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

(2)

Expert Solution

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Explanation of Solution

Given:

τ=rmg

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 2

Formula used:

τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

Calculation:

From part 1, the net torque is τ=rmg

From the rotational equation of motion:

τ=Iα

α=dωdt

On comparing both the equations:

τ=rmg=Iα

α=rmgI

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is τ=Iα and the net force is mg .

Conclusion:

Hence, the angular acceleration is α=rmgI . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

(3)

Expert Solution

To determine

The effect of friction on the torque and force.

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

τ=r×F−frictional force

Fnet=mg−frictional force

Calculation:

τ=r×F=r×mgτ=rmgsinθ−μFfτ=rmg−μFf

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.

Answer 8

(1)

Expert Solution

To determine

To Predict: The motion of the board.

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 1

Formula used:

τ=r×F=r×mg

Calculation:

τ=r×F⇒τ=r×mg⇒τ=rmgsinθ (θ=90°)⇒τ=rmg

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

Answer 9

(1)

Expert Solution

Answer 10

(1)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To Predict: The motion of the board.

Answer 13

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Answer 14

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 1

Formula used:

τ=r×F=r×mg

Calculation:

τ=r×F⇒τ=r×mg⇒τ=rmgsinθ (θ=90°)⇒τ=rmg

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

Answer 15

(2)

Expert Solution

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Explanation of Solution

Given:

τ=rmg

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 2

Formula used:

τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

Calculation:

From part 1, the net torque is τ=rmg

From the rotational equation of motion:

τ=Iα

α=dωdt

On comparing both the equations:

τ=rmg=Iα

α=rmgI

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is τ=Iα and the net force is mg .

Conclusion:

Hence, the angular acceleration is α=rmgI . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

Answer 16

(2)

Expert Solution

Answer 17

(2)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Answer 20

Explanation of Solution

Given:

τ=rmg

Tutorials in Introductory Physics, Chapter 4.3, Problem 1aT , additional homework tip 2

Formula used:

τ=Iα

Where, I is the moment of inertia and α is the angular acceleration.

Calculation:

From part 1, the net torque is τ=rmg

From the rotational equation of motion:

τ=Iα

α=dωdt

On comparing both the equations:

τ=rmg=Iα

α=rmgI

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is τ=Iα and the net force is mg .

Conclusion:

Hence, the angular acceleration is α=rmgI . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

Answer 21

(3)

Expert Solution

To determine

The effect of friction on the torque and force.

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

τ=r×F−frictional force

Fnet=mg−frictional force

Calculation:

τ=r×F=r×mgτ=rmgsinθ−μFfτ=rmg−μFf

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.

Answer 22

(3)

Expert Solution

Answer 23

(3)

Expert Solution

Answer 24

Expert Solution

Answer 25

To determine

The effect of friction on the torque and force.

Answer 26

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Answer 27

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

τ=r×F−frictional force

Fnet=mg−frictional force

Calculation:

τ=r×F=r×mgτ=rmgsinθ−μFfτ=rmg−μFf

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.