Rotational Motion

Question

If a 2.5 kg weight is hung from the 0 cm mark at one end of a meter stick that weighs 0.5 kg, where should you put your finger to balance the stick?

Answer 1

Hi Cathy L.,

You are doing a problem here involving torques.

A torque is a combination (product) of a force and a distance from an axis (= fulcrum, =pivot point, and so on).

So to do this problem, you need to balance the torque on each side of the finger-balance-point.

Torques add on each side.

So, designate the place you put your finger as distance x in cm.

Now, on the left, you have the weight (2.5.kg) at a distance x . The product of just that is T = 2.5x kg cm . But you also have the contribution of the portion of the meterstick on the left of your finger. That may be itemized as its mass, times the distance where its center of gravity lies, right in the middle of that portion. So that would be: T = (0.5 kg/100cm) * x cm * (x/2 cm) : the 0.5 kg/100 cm is the mass per unit length, the mass of that portion is the mass per unit length times the length, and the x/2 cm is the location of its center-of-mass.

So, for the torque on the left, it sums to: T = 2.5x + (0.25/100) x^2 kg cm

On the right, you have only the gravity acting on the remaining portion of the meterstick:

T = (100-x) cm * (0.5 kg/100 cm) * (100-x)/2 cm ; these are respectively the length of the right-hand portion, the mass-rate of the stick, and the position of the right-hand-portion center-of-mass from the pivot point.

That will be T = (x^2 - 200x +10000)*(0.0025) kg cm . You have to multiply this out, then equate it to the left-hand-part torque, and solve that equation for x . (It will be a quadratic equation). I'll leave the details to you!

-- Cheers, -- Mr. d.