Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 10, Problem 40AP

(a)

To determine

The total torques due to the weight of the hand about the axis of rotation when the time reads 3:00 , 5:15 , 6:00 , 8:20 and 9:45 .

(a)

Expert Solution
Check Mark

Answer to Problem 40AP

The total torques due to the weight of the hand about the axis of rotation when the time reads 3:00 , 5:15 , 6:00 , 8:20 and 9:45 are 794Nm , 2510Nm , 0 , 1160Nm and 2940Nm respectively.

Explanation of Solution

Given information: The mass of hour hand is 60kg , mass of minute hand is 100kg , length of hour hand is 2.70m , length of minute hand is 4.50m .

Formula to calculate the net torque produced by the clock’s hand is,

τ=mhg(Lh2)sinθh+mmg(Lm2)sinθm=g2(mhLhsinθh+mmLmsinθm) (1)

Here,

τ is the net torque produce by both hands.

mh is the mass of the hour hand.

mm is the mass of the minute hand.

Lh is the mass of the hour hand.

Lm is the mass of the minute hand.

θh is the angle made by the hour hand at time t .

θm is the angle made by the minute hand at time t .

Formula to calculate the angular speed of the hour hands is,

ωh=OnerevolutionTimetakenbythehourhandtocompleteonerevolution

Substitute 12h for timetakenbythehourhandtocompleteonerevolution in above equation.

ωh=1rev12h×2πrad1rev=π6rad/h

Thus, the angular speed of the hour hand is π6rad/h .

Formula to calculate the angular speed of the minute hands is,

ωm=OnerevolutionTimetakenbytheminutehandtocompleteonerevolution

Substitute 1h for timetakenbytheminutehandtocompleteonerevolution in above equation.

ωm=1rev1h×2πrad1rev=2πrad/h

Thus, the angular speed of the hour hand is 2πrad/h .

Let take t=0 at 12:00 clock.

Write the expression for the angular position of the hour hands at time t is,

θh=ωht

Here,

ωh is the angular speed of the hour hand.

Substitute π6rad/h for ωh in above equation.

θh=(π6rad/h)t

Write the expression for the angular position of the minute hands at time t is,

θm=ωmt

Here,

ωm is the angular speed of the minute hand.

Substitute 2πrad/h for ωm in above equation.

θm=(2πrad/h)t

Substitute (π6rad/h)t for θh and (2πrad/h)t for θm in equation (1).

τ=g2(mhLhsin(πt6)+mmLmsin(2πt))

Substitute 9.8m/s2 for g , 60kg for mh , 100kg for mm , 2.70m for Lh and 4.50m for Lm to fine the equation for τ .

τ=9.8m/s22((60kg)(2.70m)sin(πt6)+(100kg)(4.50m)sin(2πt))=4.9m/s2((162kgm)sin(πt6)+(450kgm)sin(2πt))=(794Nm)(sin(πt6)+2.78sin(2πt)) (2)

When clock shows time 3:00 that gives time t=3h

Substitute 3h for t in equation (2).

τ=(794Nm)(sin(π×36)+2.78sin(2π×3))=(794Nm)(sin(π2)+2.78sin(6π))=794Nm

Thus, the net torque is 794Nm .

When clock shows time 5:15 that gives time

t=(5+1560)h=5.25h

Substitute 5.25h for t in equation (2).

τ=(794Nm)(sin(π×5.25h6)+2.78sin(2π×5.25h))=(794Nm)(sin(5.25π2)+2.78sin(10.5π))=2510Nm

Thus, the net torque is 2510Nm .

When clock shows time 6:00 that gives time

t=6h

Substitute 6h for t in equation (2).

τ=(794Nm)(sin(π×66)+2.78sin(2π×6))=(794Nm)(sin(π)+2.78sin(12π))=0

Thus, the net torque is 0 .

When clock shows time 8:20 that gives time

t=(8+2060)h=8.33h

Substitute 8.33h for t in equation (2).

τ=(794Nm)(sin(π×8.33h6)+2.78sin(2π×8.33h))=(794Nm)(sin(8.33hπ6)+2.78sin(16.6π))=1160Nm

Thus, the net torque is 1160Nm .

When clock shows time 9:45 that gives time

t=(9+4560)h=9.75h

Substitute 9.75h for t in equation (2).

τ=(794Nm)(sin(π×9.75h6)+2.78sin(2π×9.75h))=(794Nm)(sin(9.75hπ6)+2.78sin(19.5π))=2940Nm

Thus, the net torque is 2940Nm .

Conclusion:

Therefore, the total torques due to the weight of the hand about the axis of rotation when the time reads 3:00 , 5:15 , 6:00 , 8:20 and 9:45 are 794Nm , 2510Nm , 0 , 1160Nm and 2940Nm respectively.

(b)

To determine

The all the time nearest to second when total torque about the axis of rotation is zero by solving the transcendental equation.

(b)

Expert Solution
Check Mark

Answer to Problem 40AP

The time corresponding to the zero torque is given as:

Time(hr) Clock time
0 12:00:00
0.515 12:30:55
0.971 12:58:19
1.54 1:32:31
1.95 1:57:01
2.56 2:33:25
2.94 2:56:29

Explanation of Solution

Given information: The mass of hour hand is 60kg , mass of minute hand is 100kg , length of hour hand is 2.70m , length of minute hand is 4.50m .

From equation (2), the expression for the total torque is given by,

τ=(794Nm)(sin(πt6)+2.78sin(2πt))

Substitute 0 for τ to find all the times.

(794Nm)(sin(πt6)+2.78sin(2πt))=0sin(πt6)+2.78sin(2πt)=0

Since it is a transcendental equation, solving the equation numerically the values of time comes out to be 0, 0.515, 0.971, 1.54, 1.95……so on.

The time corresponding to the time is given as:

Time(hr) Clock time
0 12:00:00
0.515 12:30:55
0.971 12:58:19
1.54 1:32:31
1.95 1:57:01
2.56 2:33:25
2.94 2:56:29

Conclusion:

Therefore, time corresponding to the zero torque is given as:

Time(hr) Clock time
0 12:00:00
0.515 12:30:55
0.971 12:58:19
1.54 1:32:31
1.95 1:57:01
2.56 2:33:25
2.94 2:56:29

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Chapter 10 Solutions

Physics for Scientists and Engineers with Modern Physics

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