torque question

[happyfellow]

A particle of mass m is attached to a rod of negligible mass and length L. The rod is attached to the ground at point O. If the rod makes an angle theta with horizontal as it falls, calculate the torque about point O.

A. Lmgsin(theta)
B. Lmgcos(theta)
C. mgsin(theta)
D. mgcos(theta)

I'm able to eliminate C and D because they don't take into account the length of the rod. So I'm down to A and B. I recall that the torque = rFsin(theta) so I pick A, but the answer is B. Can anyone explain?

-happy

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[lorenzomicron]

Well, the force which causes the torque is the gravity acting on the end of the rod.
we know that torque is composed of force and a perpendicular distance (length of the torque arm). thus, the length component of the rod which matters is its horizontal component.

this Lx= Lcos(theta)

then we calculate the torque: (F)(Lx)= mgLcos(theta)

hope that helps!

 

[puravida85]

I looked at this one differently..

The weight of the mass is the force acting along L (your torque). Therefore, you must break the component of the weight into its x and y components. From there, you'll look for the component that moves perpendicular to the lever arm, in this case it is cosine theta. Therefore, that is the force you are interested in using for Torque= Length * Force.

Torque = L * mg *cos(theta).

The trick is to try to place your lever arm exactly horizontal, that way you can just draw out the x and y planes much more easily and decide which force is perpendicular to the lever arm.

 

[plsfoldthx]

how can the rod fall if it's attached to the ground

 

[lorenzomicron]

The rod's end is attached. not the whole rod. since it's not a point object (i.e. existing at one point), the force can cause the rod to decline. however, in the case of torque, the rod will simply spin around the z-axis.