1.) Big Ben, the nickname for the clock in Elizabeth Tower (named after the Queen in 2012) in London, has an hour hand2.70 mlong with a mass of60.0 kgand a minute hand4.50 mlong with a mass of100 kg(figure). Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long, thin rods rotated about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per12 hoursand60 minutes,respectively.) answer in J2.)In the figure below, the hanging object has a mass ofm1=0.390kg;the sliding block has a mass ofm2=0.890kg;and the pulley is a hollow cylinder with a mass ofM= 0.350 kg,an inner radius ofR1= 0.020 0 m,and an outer radius ofR2= 0.030 0 m.Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface isk= 0.250.The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity ofvi= 0.820 m/stoward the pulley when it passes a reference point on the table.(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. answer in m/s(b) Find the angular speed of the pulley at the same moment. answer in rad/s3.)A marble and a cube are placed at the top of a ramp. Starting from rest at the same height, the marble rolls without slipping and the cube slides (no kinetic friction) down the ramp. Determine the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp.vcubevmarble=

Introduction:

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This article discusses three different physics problems related to rotational kinetic energy, energy methods, and center-of-mass speed. The first problem involves calculating the total rotational kinetic energy of the minute and hour hands of the Big Ben clock tower in London. The second problem requires using energy methods to predict the velocity of a sliding block on a pulley system. Finally, the third problem explores the difference in the center-of-mass speed between a marble and a cube rolling down a ramp.

Description:

Problem 1: The first problem focuses on the famous Big Ben clock tower located in London. The article provides information regarding the length and mass of both the hour and minute hands of the clock, and asks the reader to calculate the total rotational kinetic energy of the hands. A model of the hands as thin rods rotated about one end is used, and it is assumed that the hands are rotating at a constant rate of one revolution per 12 hours and 60 minutes, respectively.

Problem 2: In the second problem, the article discusses a pulley system comprised of a hanging object, a sliding block, and a pulley without friction on its axle. The article asks readers to predict the velocity of the sliding block using energy methods after it has moved to a second point, as well as to calculate the angular speed of the pulley at the same moment.

Problem 3: Finally, the third problem examines the differences in center-of-mass speed between a marble and a cube when rolling down a ramp. Both objects start at the same height, and it is assumed that the marble rolls without slipping while the cube slides without friction. Readers are asked to determine the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp.

Objectives:

1. To understand rotational kinetic energy and its calculation for long, thin rods.

2. To apply energy methods in predicting the speed of an object and the angular speed of a pulley.

Learning Outcomes:

1. Students will be able to calculate the total rotational kinetic energy of long, thin rods in constant rotation.

2. Students will be able to use energy methods to predict the final velocity of a moving object and the angular speed of a pulley.

3. Students will be able to compare the center-of-mass speed ratios of objects with different motion types, such as rolling and sliding.

Heading One: Rotational Kinetic Energy

1. Students will learn the concept of rotational kinetic energy and its calculation for long, thin rods.

2. Students will apply the concept to calculate the total rotational kinetic energy of the hour and minute hands of Big Ben.

3. Students will be able to model long, thin rods in constant rotation and calculate their rotational kinetic energy.

Heading Two: Energy Methods in Motion Prediction

1. Students will learn how to apply energy methods in predicting the final velocity of a moving object.

2. Students will learn how to use energy methods to calculate the angular speed of a pulley.

3. Students will apply energy methods to predict the speed of the sliding block and the angular speed of the pulley in the given scenario.

Heading Three: Comparison of Motion Types

1. Students will compare the center-of-mass speed ratios of objects with different motion types, such as rolling and sliding.

2. Students will compare the speeds of the cube and the marble at the bottom of the ramp.

3. Students will be able to understand the differences in center-of-mass speed ratios between rolling and sliding objects.

Solution 1:

To calculate the total rotational kinetic energy of the hour and minute hands of Big Ben, we need to use the formula KE(rotational) = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. Assuming both hands are long, thin rods rotated about one end and rotating at a constant rate of one revolution per 12 hours and 60 minutes respectively:

For the hour hand:

I = (1/3)ML^2,

where M is the mass of the hand and L is the length of the hand.

I = (1/3)(60.0 kg)(2.70 m)^2 = 517.20 kg m^2 (moment of inertia)

ω = 2π/12 hours = π/6 radians per hour = π/6 ÷ 3600 seconds per second = 0.00014544 radians per second (angular velocity)

KE(rotational) = (1/2)(517.20 kg m^2)(0.00014544 radians per second)^2 = 0.002278 J

For the minute hand:

I = (1/3)ML^2,

where M is the mass of the hand and L is the length of the hand.

I = (1/3)(100 kg)(4.50 m)^2 = 607.50 kg m^2 (moment of inertia)

ω = 2π/60 minutes = π/30 radians per minute = π/30 ÷ 60 seconds per second = 0.00174533 radians per second (angular velocity)

KE(rotational) = (1/2)(607.50 kg m^2)(0.00174533 radians per second)^2 = 0.008475 J

Total rotational kinetic energy of the two hands about the axis of rotation is:

KE(rotational total) = 0.002278 J + 0.008475 J = 0.010753 J

Solution 2:

To predict the speed of the sliding block after it has moved to a second point, 0.700 m away, and the angular speed of the pulley at the same moment, we need to use the principle of conservation of energy, where the initial energy is equal to the final energy.

Initial energy:

Ei = KE(translational) of the block + KE(rotational) of the pulley

KE(translational) = (1/2)m2vi^2,

where m2 is the mass of the block and vi is its initial velocity.

KE(translational) = (1/2)(0.890 kg)(0.820 m/s)^2 = 0.303590 J

KE(rotational) = (1/2)Iω^2,

where I is the moment of inertia of the pulley and ω is its angular velocity.

I = (1/2)M(R1^2 + R2^2),

where M is the mass of the pulley and R1 and R2 are the inner and outer radii of the pulley respectively.

I = (1/2)(0.350 kg)((0.020 0 m)^2 + (0.030 0 m)^2) = 0.00001050 kg m^2 (moment of inertia)

ω = vf/R2,

where vf is the final velocity of the block and R2 is the outer radius of the pulley.

We can’t solve for vf and ω individually, so we need another equation.

Final energy:

Ef = KE(translational) of the block + KE(rotational) of the pulley + KE(translational) of the pulley

KE(translational) of the pulley = (1/2)(M + m1)vf^2,

where M is the mass of the pulley, m1 is the mass of the hanging object, and vf is the final velocity of the block.

KE(rotational) of the pulley is still (1/2)Iω^2.

Equate Ei to Ef:

0.303590 J + (1/2)(0.00001050 kg m^2)(vf/R2)^2 = (1/2)(0.00001050 kg m^2)(vf/R2)^2 + (1/2)(0.350 kg + 0.390 kg)vf^2 + (1/2)Iω^2

Simplify and solve for vf:

vf = 0.6588 m/s

To find the angular speed of the pulley at the same moment:

ω = vf/R2

ω = 0.6588 m/s ÷ 0.030 0 m = 21.96 rad/s

Therefore, the predicted speed of the sliding block after it has moved to a second point, 0.700 m away, is 0.6588 m/s and the angular speed of the pulley at the same moment is 21.96 rad/s.

Solution 3:

Let vcm,cube be the center-of-mass speed of the cube and vcm,marble be the center-of-mass speed of the marble.

For the cube:

Initial energy = mgh, where m is the mass of the cube, g is acceleration due to gravity, and h is the height.

Final energy = (1/2)mvcm,cube^2

For the marble:

Initial energy = mgh, where m is the mass of the marble, g is acceleration due to gravity, and h is the height.

Final energy = (1/2)mvcm,marble^2 + (1/2)Iω^2, where I is the moment of inertia of the marble and ω is its angular velocity.

Since the height and mass are the same for both cube and marble, we can equate their initial energies:

mgh = mgh

To find the final speed of the cube and the marble, we equate their final energies:

(1/2)mvcm,cube^2 = (1/2)mvcm,marble^2 + (1/2)Iω^2

We can simplify the equation by dividing by (1/2)m:

vcm,cube^2 = vcm,marble^2 + Iω^2/m

The moment of inertia of a solid sphere is (2/5)mr^2, and the moment of inertia of a cube is (1/6)ma^2, where m is the mass and r is the radius or side length respectively. Since the problem does not specify the size of the objects, we cannot solve for their exact moment of inertia.

However, we can make a simplifying assumption that the moment of inertia of the marble is negligible compared to its translational kinetic energy, such that Iω^2/m ≈ 0. This assumption is reasonable since the marble is rolling without slipping, which means its rotational kinetic energy is converted into translational kinetic energy.

With this assumption, we can simplify the equation:

vcm,cube^2 = vcm,marble^2

Taking the square root of both sides, we get:

vcm,cube/vcm,marble = 1

Therefore, the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp is 1:1 or vcm,cube/vcm,marble = 1.

Suggested Resources/Books:

1. “University Physics with Modern Physics” by Hugh D. Young and Roger A. Freedman

2. “Fundamentals of Physics” by David Halliday, Robert Resnick, and Jearl Walker

3. “Physics for Scientists and Engineers” by Randall D. Knight

Similar Asked Questions:

1. What is the rotational kinetic energy of a system with multiple rotating objects?

2. How do you calculate the speed of an object using energy methods?

3. What is the ratio of kinetic energy to potential energy in a system with rotating objects?

4. How do you calculate the angular velocity of a pulley system?

5. What is the difference between sliding and rolling friction in terms of energy conservation?1.) Big Ben, the nickname for the clock in Elizabeth Tower (named after the Queen in 2012) in London, has an hour hand2.70 mlong with a mass of60.0 kgand a minute hand4.50 mlong with a mass of100 kg(figure). Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long, thin rods rotated about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per12 hoursand60 minutes,respectively.) answer in J2.)In the figure below, the hanging object has a mass ofm1=0.390kg;the sliding block has a mass ofm2=0.890kg;and the pulley is a hollow cylinder with a mass ofM= 0.350 kg,an inner radius ofR1= 0.020 0 m,and an outer radius ofR2= 0.030 0 m.Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface isk= 0.250.The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity ofvi= 0.820 m/stoward the pulley when it passes a reference point on the table.(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. answer in m/s(b) Find the angular speed of the pulley at the same moment. answer in rad/s3.)A marble and a cube are placed at the top of a ramp. Starting from rest at the same height, the marble rolls without slipping and the cube slides (no kinetic friction) down the ramp. Determine the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp.vcubevmarble=

Introduction:

This article discusses three different physics problems related to rotational kinetic energy, energy methods, and center-of-mass speed. The first problem involves calculating the total rotational kinetic energy of the minute and hour hands of the Big Ben clock tower in London. The second problem requires using energy methods to predict the velocity of a sliding block on a pulley system. Finally, the third problem explores the difference in the center-of-mass speed between a marble and a cube rolling down a ramp.

Description:

Problem 1: The first problem focuses on the famous Big Ben clock tower located in London. The article provides information regarding the length and mass of both the hour and minute hands of the clock, and asks the reader to calculate the total rotational kinetic energy of the hands. A model of the hands as thin rods rotated about one end is used, and it is assumed that the hands are rotating at a constant rate of one revolution per 12 hours and 60 minutes, respectively.

Problem 2: In the second problem, the article discusses a pulley system comprised of a hanging object, a sliding block, and a pulley without friction on its axle. The article asks readers to predict the velocity of the sliding block using energy methods after it has moved to a second point, as well as to calculate the angular speed of the pulley at the same moment.

Problem 3: Finally, the third problem examines the differences in center-of-mass speed between a marble and a cube when rolling down a ramp. Both objects start at the same height, and it is assumed that the marble rolls without slipping while the cube slides without friction. Readers are asked to determine the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp.

Objectives:

1. To understand rotational kinetic energy and its calculation for long, thin rods.

2. To apply energy methods in predicting the speed of an object and the angular speed of a pulley.

Learning Outcomes:

1. Students will be able to calculate the total rotational kinetic energy of long, thin rods in constant rotation.

2. Students will be able to use energy methods to predict the final velocity of a moving object and the angular speed of a pulley.

3. Students will be able to compare the center-of-mass speed ratios of objects with different motion types, such as rolling and sliding.

Heading One: Rotational Kinetic Energy

1. Students will learn the concept of rotational kinetic energy and its calculation for long, thin rods.

2. Students will apply the concept to calculate the total rotational kinetic energy of the hour and minute hands of Big Ben.

3. Students will be able to model long, thin rods in constant rotation and calculate their rotational kinetic energy.

Heading Two: Energy Methods in Motion Prediction

1. Students will learn how to apply energy methods in predicting the final velocity of a moving object.

2. Students will learn how to use energy methods to calculate the angular speed of a pulley.

3. Students will apply energy methods to predict the speed of the sliding block and the angular speed of the pulley in the given scenario.

Heading Three: Comparison of Motion Types

1. Students will compare the center-of-mass speed ratios of objects with different motion types, such as rolling and sliding.

2. Students will compare the speeds of the cube and the marble at the bottom of the ramp.

3. Students will be able to understand the differences in center-of-mass speed ratios between rolling and sliding objects.

Solution 1:

To calculate the total rotational kinetic energy of the hour and minute hands of Big Ben, we need to use the formula KE(rotational) = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. Assuming both hands are long, thin rods rotated about one end and rotating at a constant rate of one revolution per 12 hours and 60 minutes respectively:

For the hour hand:

I = (1/3)ML^2,

where M is the mass of the hand and L is the length of the hand.

I = (1/3)(60.0 kg)(2.70 m)^2 = 517.20 kg m^2 (moment of inertia)

ω = 2π/12 hours = π/6 radians per hour = π/6 ÷ 3600 seconds per second = 0.00014544 radians per second (angular velocity)

KE(rotational) = (1/2)(517.20 kg m^2)(0.00014544 radians per second)^2 = 0.002278 J

For the minute hand:

I = (1/3)ML^2,

where M is the mass of the hand and L is the length of the hand.

I = (1/3)(100 kg)(4.50 m)^2 = 607.50 kg m^2 (moment of inertia)

ω = 2π/60 minutes = π/30 radians per minute = π/30 ÷ 60 seconds per second = 0.00174533 radians per second (angular velocity)

KE(rotational) = (1/2)(607.50 kg m^2)(0.00174533 radians per second)^2 = 0.008475 J

Total rotational kinetic energy of the two hands about the axis of rotation is:

KE(rotational total) = 0.002278 J + 0.008475 J = 0.010753 J

Solution 2:

To predict the speed of the sliding block after it has moved to a second point, 0.700 m away, and the angular speed of the pulley at the same moment, we need to use the principle of conservation of energy, where the initial energy is equal to the final energy.

Initial energy:

Ei = KE(translational) of the block + KE(rotational) of the pulley

KE(translational) = (1/2)m2vi^2,

where m2 is the mass of the block and vi is its initial velocity.

KE(translational) = (1/2)(0.890 kg)(0.820 m/s)^2 = 0.303590 J

KE(rotational) = (1/2)Iω^2,

where I is the moment of inertia of the pulley and ω is its angular velocity.

I = (1/2)M(R1^2 + R2^2),

where M is the mass of the pulley and R1 and R2 are the inner and outer radii of the pulley respectively.

I = (1/2)(0.350 kg)((0.020 0 m)^2 + (0.030 0 m)^2) = 0.00001050 kg m^2 (moment of inertia)

ω = vf/R2,

where vf is the final velocity of the block and R2 is the outer radius of the pulley.

We can’t solve for vf and ω individually, so we need another equation.

Final energy:

Ef = KE(translational) of the block + KE(rotational) of the pulley + KE(translational) of the pulley

KE(translational) of the pulley = (1/2)(M + m1)vf^2,

where M is the mass of the pulley, m1 is the mass of the hanging object, and vf is the final velocity of the block.

KE(rotational) of the pulley is still (1/2)Iω^2.

Equate Ei to Ef:

0.303590 J + (1/2)(0.00001050 kg m^2)(vf/R2)^2 = (1/2)(0.00001050 kg m^2)(vf/R2)^2 + (1/2)(0.350 kg + 0.390 kg)vf^2 + (1/2)Iω^2

Simplify and solve for vf:

vf = 0.6588 m/s

To find the angular speed of the pulley at the same moment:

ω = vf/R2

ω = 0.6588 m/s ÷ 0.030 0 m = 21.96 rad/s

Therefore, the predicted speed of the sliding block after it has moved to a second point, 0.700 m away, is 0.6588 m/s and the angular speed of the pulley at the same moment is 21.96 rad/s.

Solution 3:

Let vcm,cube be the center-of-mass speed of the cube and vcm,marble be the center-of-mass speed of the marble.

For the cube:

Initial energy = mgh, where m is the mass of the cube, g is acceleration due to gravity, and h is the height.

Final energy = (1/2)mvcm,cube^2

For the marble:

Initial energy = mgh, where m is the mass of the marble, g is acceleration due to gravity, and h is the height.

Final energy = (1/2)mvcm,marble^2 + (1/2)Iω^2, where I is the moment of inertia of the marble and ω is its angular velocity.

Since the height and mass are the same for both cube and marble, we can equate their initial energies:

mgh = mgh

To find the final speed of the cube and the marble, we equate their final energies:

(1/2)mvcm,cube^2 = (1/2)mvcm,marble^2 + (1/2)Iω^2

We can simplify the equation by dividing by (1/2)m:

vcm,cube^2 = vcm,marble^2 + Iω^2/m

The moment of inertia of a solid sphere is (2/5)mr^2, and the moment of inertia of a cube is (1/6)ma^2, where m is the mass and r is the radius or side length respectively. Since the problem does not specify the size of the objects, we cannot solve for their exact moment of inertia.

However, we can make a simplifying assumption that the moment of inertia of the marble is negligible compared to its translational kinetic energy, such that Iω^2/m ≈ 0. This assumption is reasonable since the marble is rolling without slipping, which means its rotational kinetic energy is converted into translational kinetic energy.

With this assumption, we can simplify the equation:

vcm,cube^2 = vcm,marble^2

Taking the square root of both sides, we get:

vcm,cube/vcm,marble = 1

Therefore, the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp is 1:1 or vcm,cube/vcm,marble = 1.

Suggested Resources/Books:

1. “University Physics with Modern Physics” by Hugh D. Young and Roger A. Freedman

2. “Fundamentals of Physics” by David Halliday, Robert Resnick, and Jearl Walker

3. “Physics for Scientists and Engineers” by Randall D. Knight

Similar Asked Questions:

1. What is the rotational kinetic energy of a system with multiple rotating objects?

2. How do you calculate the speed of an object using energy methods?

3. What is the ratio of kinetic energy to potential energy in a system with rotating objects?

4. How do you calculate the angular velocity of a pulley system?

5. What is the difference between sliding and rolling friction in terms of energy conservation?

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