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.
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?
2 answers
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
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
0 answers