Flagpole problem concerning torque.

[M1ZeN]

Homework Statement

A flag pole consists of two parts. The first part is the rod, which has a mass of 20 kg and a length of 5 m. The second part is a solid ball at one end, which has a mass of 0.5 kg. If the flag pole comes out of the building at an angle of 90 degrees (perpendicular to the building), what is the magnitude of the torque exerted by the entire flagpole at the point of contact with the building? Assume that the solid ball is on the end of the rod opposite of the building.

Homework Equations

(vector) T = (vector) r X (vector) F
(vector) T = I * angular acceleration
Tgrav = -MgXcm, Xcm is the center of mass

The Attempt at a Solution

I did an attempt at the problem with the equation, Tgrav = -MgXcm.

I considered the center of mass would be at the middle of the rod, 5 m / 2 = 2.5 m. The collective mass to compute into the equation would be combined from 20 kg (the rod) with the 0.5 kg (solid ball).

Tgrav = -MgXcm
Tgrav = -(20.5 kg)(9.8 m/s^2)(2.5m)
Tgrav = -502.3 Nm

This is the only way I found to use all the values that are presented in the problem. Since the pivot point is the end of the rod attached to the building, the rod is essentially falling towards the Earth clockwise thus making the torque negative. The way I approached seems reasonable, but part of me is not sure.

Any suggestions?

Thanks

 

[denverdoc]

Not a bad effort. But what is a bit amiss is the assumption that the cg of the system will remain in the center when you add the contribution of the ball. Imagine a seesaw--i you add weight to one end, it will tip as the cg is not centered.

You have two choices: you can just add the torques separately which is the most straightforward, or you can compute the new cg of the rod/ball combination and use that value of x in conjunction with the combined mass. The math will end up looking identical.

If you went with the second approach: you'd have (all measurements relative to the bldg)

cg=[20kg(2.5m)+0.5Kg(5)]/(20+0.5)

=52.5/20.5=2.56