Question

Problem 10.32

A very thin horizontal, 2.00-m long, 5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. At the east end of the beam, a 200-N force pushes downward. At the west end of the beam, a 200-N force pushes upward. What is the angular acceleration of the beam?

A: 1.33 × 102 rad/s2 south
B: 240 rad/s2 south
C: zero
D: 240 rad/s2 north
E: 1.33 × 102 rad/s2 north

I know the answer is supposed to be D, 240 rad/s2 north; however I am wrecking my mind trying to figure out the logic in that. I would appreciate an explanation on that :)

Answer 1

Torque is given by:\[\tau =I \alpha\]where τ is torque; I is the moment of inertia; and alpha is the angular acceleration. We can also write this as:\[rF=I \alpha \]You know the forces acting on the beam, and you know the distance from where those forces act to the center of the beam. The moment of inertia of a thin beam rotating about its center is the same as that for a thin rod:\[I=\frac{ 1 }{ 12 }mL ^{2}\]where m is the mass of the rod; and L is the length of the rod.
We also know that torque is given by a cross-product:\[\tau =r \times F\]That means you can use the right hand rule to determine the orientation of the torque vector.

Answer 2

I have gotten that far, but I have been unable to determine that it's angular acceleration is "north". I can vaguely remember that there was a "right hand rule"; is that the key that determine the direction (north)? Thanks for the informative reply :)

Answer 3

With your right hand, curl your fingers in the direction of the beam rotation. Your thumb will point in the direction of the angular acceleration vector. Admittedly the choice of directions combine poorly with their description of the problem.

Answer 4

Thanks a lot. Much appreciated.