Solving for Angular Velocity and Torque: Big Ben Minute and Hour Hands

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SUMMARY

The discussion focuses on calculating the angular velocity, moment of inertia, rotational kinetic energy, and torque for the minute and hour hands of Big Ben. The minute hand, made of copper, has a length of 4.27 meters and a mass of 101.6 kg, while the hour hand, made of gunmetal, has a length of 2.74 meters and a mass of 304.8 kg. Key equations include angular velocity as delta theta over delta time, moment of inertia as mr squared, and torque calculated using the formula Torque = R x F. The torque varies based on the position of the hands, being zero when vertical and maximum when horizontal.

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HELP! I am desperate!

1. Homework Statement
The minute hand of big ben is made of copper, has a length of 4.27 meters, and a mass of 101.6kg. The hour hand is made of gunmetal, has a length of 2.74m, and a mass of 304.8kg. Assume that each hand is a uniform rod and rotates about one end. Use a cartesian coordinate system with +x to the right, +y up, and +z out of page. Find the angular velocity, the moment of inertis, the rotational KE, and the angular momentum of the minute and hour hand. Then calculate the torque of the drive shaft owing to the weight if the hand acting at its center of mass when the arm points to the numbers 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. Torque=R x F AT There should be 34 answers in all.


2. Homework Equations
I don't know where to start with this problem because I don't know which equations to use.
Angular velocity= delta theta/ delta time??
moment of inertia= mr squared ((g Tm squared/ 2h) -1)?
rotational KE= theta final-theta initial/ tfinal- tinitial?


3. The Attempt at a Solution

My attempt at a solution failed because I do not think the equations are correct. I have been trying this problem for a week and can not figure it out. I do not even know where to start from.
 
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Angular velocity looks right.
Rotational kinetic energy equals 1/2*moment of inertia*angular velocity squared
Torque = RxF. The force to consider is the force of gravity, which can be assumed to operate through a point halfway along the relevant hand. The torque will be zero when the hand is vertical and maximum when the hand is horizontal.
 

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