a mass of 100 gram is attached to the end of a rubber string 49 cm long and having an area of cross section 20 sq. mm. the string is whirled round, horizontally at a constant speed of 40 r.p.s. in a circle of radius 51 cm. find the Young's modulus of rubber. explain the answer also.

Asked by ABHINAVSINGH | 22nd Feb, 2014, 10:00: AM

Expert Answer:

Given that,
 
L = 49 cm
l = 51 - 49 = 2 cm
Cross sectional area a  20 mm2 = 0.2 cm2
m = 100 g
R = 51 cm
ω = 40 r.p.s = 80 π rad s-1
 
When the mass attached to the string is whirled, the length of the string increases under the effect of centrifugal force.
 
The centrifugal force F = mRω2
                                = 100 x 51 x (80 π)2
 

Answered by  | 24th Feb, 2014, 01:28: PM

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