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2 years ago

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find the resultant in the x-direction , Rx and y-direction , Ry for F1 = 2.5n in the positive x-direction f2 = 1n in the negativ

e x-direction f3= 2n in the positive y-direction F4=3n in the negative y-direction

1 answer:

Advocard [28]2 years ago

5 0

The resultant in the x-direction:

Rx = F1 + F2 = 2.5 N - 1 N = 1.5 N .

The resultant in the y-direction:

Ry = F3 + F4 = 2 N - 3 N = -1 N.

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A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

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where

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Substituting,

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The moment of inertia of the two rings at the beginning is

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$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

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so the moment of inertia of the rings is

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and the total moment of inertia is

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When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

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$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

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Answer:

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