The experiment you did in lab is repeated, using a uniform metal bar that is 80.

Question

The experiment you did in lab is repeated, using a uniform metal bar that is 80.0 cm long instead of the meterstick. Since the bar is uniform, its center of gravity is at its center. The new experiment uses different hooks for hanging the masses from the bar, with mhook = 5.0 g. As in the experiment you did in lab, x1 = 5.00 cm, m1 = 300.0 g, and xp = 25.0 cm. In the new experiment, you make the same measurements as in your lab and plot x versus 1 m2 + mhook . The line that is the best fit to your data has slope 2200 cm · g. What is the mass of the bar?

Explanation / Answer

Use this equation,

?P = (m1 + mhook)g(xP ? x1) ? Mg(xcg ? xP) ? (m2 + mhook)g(x ? xP).

x1 = 5.0 cm

xP = 25.0 cm

m1 = 300.0 g

mhook given

By changing the equation we can find a smaller portion of it that is equal to slope, then solve for the mass of the bar (M) by setting that portion equal to 2200.

0 = (300+5)(25-5)g - Mg(40-25) - (m2+5)(x-25)g

0 = 6100g - 15Mg - (m2+5)(x-25)g

-6100g + 15Mg = -(m2+5)(x-25)g

(-6100g + 15Mg)/(-g(m2+5)) = x-25

(-6100g + 15Mg)/(-g) * 1/(m2+5) + 25 = x

Presented like this you can see a slight resemblance to an equation for a line. The portion we will use to solve for M not only contains that variable but also most closely resembles the slope of the equation. We will set it equal to 2200 and solve.

(-6100g + 15Mg)/(-g) = 2200

M = 260 g

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