A block of mass 1 kg is fastended to one end of a copper wire of cross- sectional area 1 mm^(2) and is rotated in a horizontal circle of radius 20 cm. If the breaking stress of copper is 5 xx 10^(8)Nm^(-2), find the maximum number of revolutions the block make in the minutewithout the string breaking.

A block of mass 1 kg is fastended to one end of a copper wire of cross

1.	A block of mass 1 kg is fastended to one end of a copper wire of cross- sectional area 1 mm^(2) and is rotated in a horizontal circle of radius 20 cm. If the breaking stress of copper is 5 xx 10^(8)Nm^(-2), find the maximum number of revolutions the block make in the minutewithout the string breaking.
Answer» Solution :Maximum tension on the string = Breaking stress x Area of cross-section `= 5xx10^8 xx1xx10^(-6) = 500 N.` When a body REVOLVES in a horizontal circle, Tension on the string = Centripetal force `500 = mr omega^2` Where m= 1 kg, R = 0.20m, `omega`= ? `500= 1 xx 0.2 xx omega^2` `omega^2 = (500)/(0.2) = 2500` Maximum angular speed, `omega = 50 RAD s^(-1)` `omega = (2pi m)/(t) , 50 = (2pim )/(60)` The maximum number of revolutions, `n = (50 xx 60)/(2pi) = (1500)/(pi) = 477.4 r`pm.