Rotational Kinetic Energy of Big Ben

Click For Summary
SUMMARY

The discussion focuses on calculating the total rotational kinetic energy of the hour and minute hands of Big Ben. The hour hand, with a length of 2.70 m and mass of 300 kg, has a rotational inertia of 729 kg·m² and an angular speed of 1.45×10⁻⁴ rad/s. The minute hand, measuring 4.20 m and weighing 100 kg, has a corrected rotational inertia of 588 kg·m² and an angular speed of 1.75×10⁻³ rad/s. The final formula for rotational kinetic energy combines both hands' contributions, leading to the correct total energy calculation.

PREREQUISITES
  • Understanding of rotational dynamics and kinetic energy
  • Familiarity with the concept of rotational inertia
  • Knowledge of angular velocity and its units
  • Basic proficiency in algebra for calculations
NEXT STEPS
  • Learn about the moment of inertia for various shapes and their applications
  • Study the principles of rotational kinetic energy in physics
  • Explore the effects of angular speed on rotational dynamics
  • Investigate real-world applications of rotational energy in engineering
USEFUL FOR

Students and professionals in physics, mechanical engineering, and anyone interested in the dynamics of rotating systems, particularly in the context of large structures like clocks.

Bashyboy
Messages
1,419
Reaction score
5
The problem I am working on is:

"Big Ben, the Parliament tower clock in London, has an hour hand 2.70 m long with a mass of 300 kg, and a minute hand 4.20 m long with a mass of 100 kg (see figure below). 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 per 12 hours and 60 minutes, respectively.)"

(Converted) Angular Speed of Clock Hands:

Hour Hand 1.45⋅10^{−4} rad/s
Minute Hand 1.75⋅10^{−3} rad/s

Rotational Inertia:

Hour Hand I=1/3(300 kg)(2.70 m)^2=729 kg⋅m^2
Minute Hand I=1/3(100 kg)(4.20 m)^2=243 kg⋅m^2

Rotational Kinetic Energy:

K_{rot}=1/2(729 kg⋅m^2)(1.45⋅10^{−4} rad/s)^2+1/2(243 kg⋅m^2)(1.75⋅10^{−3} rad/s)^2

When I calculate this, it comes out incorrect, what has happened?
 
Last edited:
Physics news on Phys.org
Bashyboy said:
Minute Hand I=1/3(100 kg)(4.20 m)2=243 kg⋅m^2

Check this.
 
It worked with the corrected value! Thank you so much!

The rotational inertia of the minute hand is I=588~kg\cdot m^2
 

Similar threads

Replies
9
Views
3K
Replies
6
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
15
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K