A very light rigid rod with a length of 1.03 m extends straight out from one end

Question

A very light rigid rod with a length of 1.03 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation.

Ip = ICM + MD2

(a) Determine the period of oscillation. [Suggestion: Use the parallel-axis theorem equation given above. Where D is the distance from the center-of-mass axis to the parallel axis and M is the total mass of the object.]

_______________ s

(b) By what percentage does the period differ from the period of a simple pendulum 1 m long?

Explanation / Answer

IP=ML2/12 +MD2 =M*12/12 +M*(1.03+(1/2))2

IP=2.424233M

Period of oscillation

T=2pisqrt(I/MgD] =2pi*sqimrt[2.424233M/M*9.81*(1.03+0.5)]

T=2.525 s

b)

Period With L=1 m

T=2pi*sqrt[1/9.81] =2.006 s

% difference =[(2.525-2.006)/2.006]*100

=25.87 %

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