A particular tower clock similar to Big Ben has an hour hand 2.70 m long with a mass of 62.0 kg, and a minute hand 4.80 m long with a mass of 100 kg. Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long thin rods.)
To calculate the total rotational kinetic energy of the two hands, we need to use the formula for rotational kinetic energy:
Kinetic Energy (KE) = (1/2) I ω²
where I is the moment of inertia and ω is the angular velocity.
Let's first find the moment of inertia for each hand. For a thin rod rotating about one end, the moment of inertia is given by:
I = (1/3) m L²
where m is the mass of the rod and L is the length of the rod.
For the hour hand:
Mass (m₁) = 62.0 kg
Length (L₁) = 2.70 m
Moment of inertia (I₁) = (1/3) (62.0 kg) (2.70 m)²
For the minute hand:
Mass (m₂) = 100 kg
Length (L₂) = 4.80 m
Moment of inertia (I₂) = (1/3) (100 kg) (4.80 m)²
Now, let's calculate the angular velocity (ω). The angular velocity is given by the ratio of the rotational displacement (angle) to the time taken to traverse that displacement.
For the hour hand, the angular displacement is 2π radians (one complete revolution) and it takes 12 hours (or 43,200 seconds) for the hour hand to complete one revolution.
Angular velocity (ω₁) = (2π radians) / (43,200 seconds)
For the minute hand, the angular displacement is also 2π radians and it takes 60 minutes (or 3,600 seconds) for the minute hand to complete one revolution.
Angular velocity (ω₂) = (2π radians) / (3,600 seconds)
Now, let's substitute the values into the formula for kinetic energy:
KE₁ = (1/2) I₁ ω₁²
KE₂ = (1/2) I₂ ω₂²
Total rotational kinetic energy (KE) = KE₁ + KE₂
After plugging in the moment of inertia and angular velocity values, you can calculate the total rotational kinetic energy.