Angular Acceleration Using Work and Energy Principals

[mm391]

Homework Statement

The chain of a workshop crane is 50 m long and has a mass of 3 kg/m. It is partially wound on a drum and the effective radius from the axis of the drum to the chain centre line is 0.2 m. The drum itself, including shaft and gear wheel has a mass of 100 kg and has a radius of gyration of 0.15 m. A steel block with a mass of 500 kg is to be lifted from a point 20 m below the level of the axis of the drum. If a torque of 1.3 kNm is applied to the drum, what will be the initial vertical acceleration of the steel block?

Homework Equations

P.E = mgh
K.E = 1/2mv^2
I = mK^2
Torque = radius * Force
Work Done = F*distance
F=mass* acceleration

The Attempt at a Solution

 

[mm391]

I have just found out it is meant to be using angular and linear momentum but I a m still stuck:

Sum of moments = angular momentum

Momentum = Inertia*angular velocity + mass*velocity*radius

 

[LawrenceC]

Sum of moments is not angular momentum. Sum of moments is the moment of inertia multiplied by the angular acceleration.

 

[mm391]

I solved it many thanks for your help.

 

[DeeNos]

To calculate the angular acceleration, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done on the crane's chain and drum system is equal to the change in potential energy of the steel block.

First, we need to calculate the potential energy of the steel block at its initial position, 20 m below the drum's axis. Using the equation P.E = mgh, we get P.E = (500 kg)(9.8 m/s^2)(20 m) = 98,000 J.

Next, we need to calculate the work done on the system by the torque applied to the drum. Using the equation Work Done = F*distance, we get Work Done = (1.3 kNm)(0.2 m) = 260 J.

Since the work done is equal to the change in potential energy, we can set these two equations equal to each other and solve for the acceleration of the steel block.

260 J = 98,000 J + 1/2(500 kg)v^2

Solving for v, we get v = 17.3 m/s.

Now, we can use the equation F=mass* acceleration to calculate the force acting on the steel block. F = (500 kg)(17.3 m/s^2) = 8,650 N.

Finally, we can use the equation Torque = radius * Force to calculate the torque required to produce this force. Torque = (0.2 m)(8,650 N) = 1.73 kNm.

Therefore, the initial vertical acceleration of the steel block would be 17.3 m/s^2 and the torque required to produce this acceleration would be 1.73 kNm.