# Torque, Rod Balanced on Fulcrum

• rmunoz

## Homework Statement

The figure below shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 17 cm and L2 = 85 cm. The rod is held horizontally on the fulcrum and then released.
(a) What is the magnitude of the initial acceleration of particle 1?
(b) What is the magnitude of the initial acceleration of particle 2?

http://www.webassign.net/halliday8e/pc/halliday8019c10/halliday8019c10-fig-0042.htm [Broken]

## Homework Equations

$$\tau$$= F(perp)r = ma(tan)*r = m($$\alpha$$r)r = (mr$$^{2}$$)$$\alpha$$

$$\tau$$=$$\tau$$1 + $$\tau$$2 ?

F(tan)=mat

$$\tau$$= I$$\alpha$$

## The Attempt at a Solution

So I have no idea really what I'm doing with this problem but i figured its better to at least try something and fail than to have not tried at all... here is my attempt... its nowhere near right, I allready know that, but can someone at least point me in the right direction?
I'm beyond confused as to where i should even start but perhaps there is some validity in the stuff i put together.

first) Tnet is obviously what we are looking for in order to deduce the acceleration from the known quantities. At the instant the rod begins to turn, the right side is moving downwards and the left is moving upwards, meaning that the torque generated by the force acting on P2 is greater than that of P1.

This is where i go wrong (I'm allmost 99 percent sure i have no idea what I am doing here)

$$\tau$$net= r1f1 + r2f2 = I1$$\alpha$$1 + I2$$\alpha$$2

r1f1 = -9.8m/s$$^{2}$$*m*.17m but m is irrelevant because it is the same for both particles therefore

r1f1 = -9.8m/s$$^{2}$$*.17m
&
r2f2 = -9.8m/s$$^{2}$$*.85m

similarly, $$\tau$$= mr$$^{2}$$$$\alpha$$

I'd prefer not to go any farther because it just gets uglier and its quite time consuming using latex reference. On that note, what the hell do i do to start this problem correctly?

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first) Tnet is obviously what we are looking for in order to deduce the acceleration from the known quantities. At the instant the rod begins to turn, the right side is moving downwards and the left is moving upwards, meaning that the torque generated by the force acting on P2 is greater than that of P1.
OK. (The figure is not viewable, but I think I understand.) Realize that the torques on each mass are opposite (one clockwise, the other counterclockwise) and thus have different signs.

This is where i go wrong (I'm allmost 99 percent sure i have no idea what I am doing here)

$$\tau$$net= r1f1 + r2f2 = I1$$\alpha$$1 + I2$$\alpha$$2
Let's call the torque on m1 positive (counterclockwise), the other negative (clockwise). Realize that there is a single α. So:
τ = r1f1 - r2f2 = (I1 + I2)α

r1f1 = -9.8m/s$$^{2}$$*m*.17m but m is irrelevant because it is the same for both particles therefore
The mass is not irrelevant; keep it.

r1f1 = -9.8m/s$$^{2}$$*.17m
&
r2f2 = -9.8m/s$$^{2}$$*.85m
Recalculate the net torque, taking care with signs. You might want to do it with symbols first, then plug in numbers later on when you solve for alpha. For example: use mgr2 & mgr1 for the magnitudes of the torques.

similarly, $$\tau$$= mr$$^{2}$$$$\alpha$$
Find the total I for the system by adding the I for each mass. Yes, I = mr².