Principles of Physics
Principles of Physics
5th Edition
ISBN: 9781133712725
Author: SERWAY
Publisher: CENGAGE CO
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Chapter 10, Problem 66P

The hour hand and the minute hand of Big Ben, the Parliament tower clock in London, are 2.70 m and 4.50 m long and have masses of 60.0 kg and 100 kg, respectively (see Fig. P10.17).

  1. (a) Determine the total torque due to the weight of these hands about the axis of rotation when the time reads (i) 3:00, (ii) 5:15, (iii) 6:00, (iv) 8:20, and (v) 9:45. (You may model the hands as long, thin, uniform rods.)
  2. (b) Determine all times when the total torque about the axis of rotation is zero. Determine the times to the nearest second, solving a transcendental equation numerically.

(a)

Expert Solution
Check Mark
To determine

The torque due to the weight of the hands at time 3.00, 5.15, 6.00, 8.20, and 9.45.

Answer to Problem 66P

  1. (i) The torque due to the weight of the hands at time 3.00 is 794Nm_.
  2. (ii) The torque due to the weight of the hands at time 5.15 is 2510Nm_.
  3. (iii) The torque due to the weight of the hands at time 6.00 is 0Nm_.
  4. (iv) The torque due to the weight of the hands at time 8.20 is 1160Nm_.
  5. (v) The torque due to the weight of the hands at time 9.45 is 2940Nm_.

Explanation of Solution

Consider that the weight of each hand is concentrating at the center of gravity

Write the expression for the total torque about the center of gravity.

  τ=mhg(Lh2)sinθhmmg(Lm2)sinθm=g2(mhLhsinθh+mmLmsinθm)        (I)

Here, mh is the mass of the hour hand, mm is the minute hand, Lh is the length of the hour hand, Lm is the length of the minute hand, θm is the angle inclined by minute hand, θh is the angle inclined by hour hand.

Consider t=0 as 12 o clock, hence the angular position of the hands are θh=ωht, where ωh=π6rad/h, and θm=ωmt, where ωm=2πrad/h.

Substitute, π6t for θh, and 2πt for θm in equation (I).

  τ=g2(mhLhsin(π6t)+mmLmsin(2πt))        (II)

Conclusion:

(i) Case 1: at t=3.00h.

Substitute, 60.0kg for mh, 2.70m for Lh, 100kg for mm, 4.50m for Lm, and 3.00h for t in equation (II).

  τ=9.80m/s22((60.0kg)(2.70m)sin(π6×3.00h)+(100.0kg)(4.50m)sin(2π×3.00h))=794Nm[sin(π2)+2.78sin6π]=794Nm

(ii) Case 2: at t=5.15h.

Substitute, 60.0kg for mh, 2.70m for Lh, 100kg for mm, 4.50m for Lm, and 5h+1560h for t in equation (II).

  τ=9.80m/s22((60.0kg)(2.70m)sin(π6×(5h+1560h))+(100.0kg)(4.50m)sin(2π×(5h+1560h)))=2510Nm

(iii) Case 3: at t=6.00h.

Substitute, 60.0kg for mh, 2.70m for Lh, 100kg for mm, 4.50m for Lm, and 6.00h for t in equation (II).

  τ=9.80m/s22((60.0kg)(2.70m)sin(π6×(6.00h))+(100.0kg)(4.50m)sin(2π×(6.00h)))=0Nm

(iv) Case 4: at t=8.20h.

Substitute, 60.0kg for mh, 2.70m for Lh, 100kg for mm, 4.50m for Lm, and 8h+2060h for t in equation (II).

  τ=9.80m/s22((60.0kg)(2.70m)sin(π6×(8h+2060h))+(100.0kg)(4.50m)sin(2π×(8h+2060h)))=1160Nm

(v) Case 5: at 9.45h.

Substitute, 60.0kg for mh, 2.70m for Lh, 100kg for mm, 4.50m for Lm, and 9h+4560h for t in equation (II).

  τ=9.80m/s22((60.0kg)(2.70m)sin(π6×(9h+4560h))+(100.0kg)(4.50m)sin(2π×(9h+4560h)))=2940Nm

Therefore,

  1. (i) The torque due to the weight of the hands at time 3.00 is 794Nm_.
  2. (ii) The torque due to the weight of the hands at time 5.15 is 2510Nm_.
  3. (iii) The torque due to the weight of the hands at time 6.00 is 0Nm_.
  4. (iv) The torque due to the weight of the hands at time 8.20 is 1160Nm_.
  5. (v) The torque due to the weight of the hands at time 9.45 is 2940Nm_.

(b)

Expert Solution
Check Mark
To determine

The time at which the total toque become zero.

Answer to Problem 66P

The time at which the total toque become zero is given below.

Principles of Physics, Chapter 10, Problem 66P , additional homework tip  1

Explanation of Solution

The total torque become zero, when:

  sin(πt6)+2.78sin2πt=0        (III)

Use equation (III) to find the times at which torque become zero.

Conclusion:

The times at which torque become zero is given below

Principles of Physics, Chapter 10, Problem 66P , additional homework tip  2.

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Principles of Physics

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