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Fri December 09, 2011 By: Abhishek Saxena
a Copper wire of negligible mass 1m long and cross sectional area 0.000001m^2 is kept on a smooth horizontal table with 1 end fixed. A ball of mass 1Kg is attached to the other end. The wire and the ball are rotating with an angular velocity of 20 rad/s. If the elongation of the wire is 0.001m. Find Young Modulus of wire? If on increasing the velocity to 100 rad/s the wire breaks down. Find the breaking Stress?
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Area of crosssection = 0.000001 m^{2}
(pi) r^{2} = 0.000001
r = 0.001/sqrt(pi)
Young's Modulus is defined as (F/A)/(dL/L) where F is the force, A the cross sectional area, dL the change in length and L the original length.
In this case F is a combination of the centripetal force and the gravitational force.
F = mrw^2 +mg where w is the angular velocity
= 1 * 0.001/sqrt(pi) * (20)^{2} + 1 * 9.8
In this case F is a combination of the centripetal force and the gravitational force.
F = mrw^2 +mg where w is the angular velocity
= 1 * 0.001/sqrt(pi) * (20)^{2} + 1 * 9.8
Y = (1 * 0.001/sqrt(pi) * (20)^{2} + 1 * 9.8) / 0.000001 * (0.001)/1
Young's Modulus is a measure of the stiffness of a material. It states how much a material will stretch (i.e., how much strain it will undergo) as a result of a given amount of stress.
If we put angular velocity 100 instead of 20 in above formula , we will get the breaking stress.
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