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November 27, 2014

November 27, 2014

Posted by **Hannah** on Tuesday, June 5, 2012 at 5:07pm.

I did sqrt 2.79 = 1.67 and then 1/1.67=0.59

Is this correct?

- Physics(Please check) -
**Elena**, Tuesday, June 5, 2012 at 5:27pmLet the point where the axis passes through the plate be O.

OA =sqrt (L1²+L2²),

OB = L1

a(centripetal) = ω²•R,

a(A) =ω²•OA= ω²•sqrt (L1²+L2²),

a(B) = ω²•OB = ω²•L1.

a(A)/a(B) = ω²•sqrt(L1²+L2²)/ω²•L1 = n,

n= sqrt (L1²+L2²)/ L1,

n²=(L1²+L2²)/(L1)² =1+(L2/L1)²

(L2/L1)² = n²-1,

L2/L1 = sqrt(n²-1),

L1/L2=1/ sqrt(n²-1) = =1/sqrt(2.8²-1)=3.8.

- Physics(Please check) -
**RJ**, Tuesday, November 6, 2012 at 3:52pmThis is how the solution is reached:

Centripital acceleration = radius x (angular speed)^2

The acceleration of A = 2.8 (the acceleration of B)

So to set it up we get:

Sqrt(L2^2+L1^2) w^2 = 2.8 (L1) w^2

The angular velocities cancel because they are equal.

Now square both sides to get:

(L2^2+L1^2) = 7.84 (L1)^2

Move the L1 over:

(L2^2+L1^2)/L1^2=7.84

The fraction can be written:

L2^2/L1^2 + L1^2/L1^2 = 7.8

Simplify:

L2^2/L1^2 + 1= 7.84

L2^2/L1^2= 6.84

Take the sqrt of that:

L2/L1= 2.615

The reciprocal of that is 1/2.615

L1/L2= .382

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