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quiz23_solutions - The oscillation frequency is therefore...

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05/07/08 Physics 201 Quiz 23 A ball (mass m , moment of inertia I , radius R ) is rolling near the bottom of a bowl. The bowl is shaped as y(x) =1 cm + b x 2 where b is a constant that describes the curvature of the bowl. Find the oscillation frequency of the ball rolling near the bottom of the bowl. ANS Method 1: E=1/2(m+I/R^2)v^2+mgy=1/2(m+I/R^2) v^2+mg(1 cm +bx^2). This is analogous to the spring problem with mass M=m+I/R^2 and spring constant k=2mgb.
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Unformatted text preview: The oscillation frequency is therefore f=omega/(2 pi)=1/(2 pi)*sqrt(k/M)=1/(2 pi)*sqrt(2mgb/(m+I/R^2)). Method 2: (m+I/R^2) a=F=-mg(dy/dx)=-2 mgb x identical to a spring. Again, the effective mass is M=m+I/R^2 and the effective spring constant is 2mgb. note that the 1 cm offset in the definition of y(x) does not affect the oscillation frequency....
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