A very light rigid rod with a length of 0.80 m extends straight out from one end

Question

A very light rigid rod with a length of 0.80 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.] (b) By what percentage does the period differ from the period of a simple pendulum 1 m long?

Explanation / Answer

for this type of system

a) the angular frequency is w = sqrt(3g/2L)
and the period is T = 2pi/w
= 2pi/sqrt(3g/(2L))
= 2*pi*sqrt(2L/(3g))

For L = 0.8 m,
T = 2*3.14*(2*0.8/(3*9.8))^0.5
= 1.465 s

b) For a simple pendulum of length 1 m,
T = 2*3.14*sqrt(1/9.8)
= 2.007 s
The difference is
(2.007-1.465)/1.465 = 0.3699
= 36.99 %

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