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

[dban33]

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.

 

[joeyar]

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.

 

[thomate1]

I understand your frustration and I'm here to help. Let's break down the problem step by step.

First, we need to find the angular velocity of each hand. Angular velocity is defined as the change in angular displacement over time. We can use the equation ω=Δθ/Δt, where ω is the angular velocity, Δθ is the change in angular displacement, and Δt is the change in time. In this case, we can assume that the minute hand completes one full rotation (2π radians) in 60 minutes, and the hour hand completes one full rotation in 12 hours. This means that the angular velocities of the minute and hour hand are 2π/60 radians per minute and 2π/12 radians per hour, respectively.

Next, we need to find the moment of inertia of each hand. The moment of inertia is a measure of an object's resistance to changes in its rotational motion. For a uniform rod rotating about one end, the moment of inertia is given by I=1/3ML^2, where M is the mass of the rod and L is the length. Using this equation, we can calculate the moment of inertia for both the minute and hour hand.

Now, let's calculate the rotational kinetic energy of each hand. The rotational kinetic energy is given by KE=1/2Iω^2, where I is the moment of inertia and ω is the angular velocity. We can plug in the values we calculated earlier to find the rotational kinetic energy of each hand.

Finally, we need to calculate the torque of the drive shaft due to the weight of each hand. Torque is defined as the product of force and the perpendicular distance from the point of rotation. In this case, the force is the weight of each hand and the perpendicular distance is the length of each hand. We can use the equation τ=RFsinθ, where τ is the torque, R is the length of the hand, F is the weight of the hand, and θ is the angle between the hand and the vertical axis (which is 90 degrees in this case). We can calculate the torque for each hand when it points to the numbers 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.

I hope this helps guide you in the right direction