A 11.3-kg mass, fastened to the end of an aluminum wire with an unstretched leng

Question

A 11.3-kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.71 m , is whirled in a vertical circle with a constant angular speed of123 rev/min . The cross-sectional area of the wire is0.014 cm2 .

Part A.

Calculate the elongation of the wire when the mass is at the lowest point of the path.

Express your answer with the appropriate units.

Part B:

Calculate the elongation of the wire when the mass is at the highest point of its path.

Express your answer with the appropriate units.

Explanation / Answer

Youngs modlulus Y of a material is given Y = FL/ADL

A = area

DL = change of length

L = original length

F = force

so here we apply F at lowest point

Fd = mg + m r w^2

1 RPM = 2pi/60 rev/s

Fd = (11.3 * 9.81) + (11.3 * 0.71 *(123* 2*3.14/60) * (123 *2*3.14/60))

Fd = 1440.58 N

Y of Al = 6.9 *10^10

dL = elongation = FL/AY

dL = (1440.58 * 0.71)/(0.014 e -4 * 6.9e10)

dL = 1.05 cm is the elongation

---------------------------------------

F at highest point = mr w^2 - mg

F h = -(11.3 * 9.81) + (11.3 * 0.71 *(123* 2*3.14/60) * (123 *2*3.14/60))

Fh = 1218.878 N

dL = (1218.178 * 0.71)/(0.014 e -4 * 6.9 e 10)

dL = 0.089 cm is the elongtion

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