A very light rigid rod with a length of 1.44 m extends straight out from one end

Question

A very light rigid rod with a length of 1.44 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. WhereD 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

the period is:
T = 2pi/w
= 2pi/sqrt(3g/(2L))
= 2*pi*sqrt(2L/(3g))

For L = 1.44 m,
T = 2*pi*sqrt(1.44*2/(3*9.8))
= 1.965 s

b) For a simple pendulum of length 1 m,
T = 2*pi*sqrt(1.44*2/9.8)
= 3.4 s
The difference is
(3.4-1.965)/1.965 = 0.73
= 73%

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