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March 25, 2017

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A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner A is n times as great as that measured at corner B. What is the ratio L1/L2 of the lengths of the sides of the rectangle when n = 2.80?

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

Is this correct?

  • Physics(Please check) - ,

    Let 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) - ,

    This 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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