Mechanical Properties of Solids
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Question : 11
Total: 21
A 14.5 kg mass, fastened to the end of a steel wire of unstretchedlength 1.0 m, is whirled in a vertical circle with an angular velocity of2 rev/s at the bottom of the circle. The cross-sectional area of the wire is0.065 cm2
Solution:
Here, m = 14.5 kg;
A = 0.065 c m 2 = 0.065 × 10 − 4 m 2 ;
L = 1 m v = 2 r . p . s .
When the mass is at the lowest point of its circular path, the stretching force on the wire,
F = m g + m r ω 2
= m g + m r × ( 2 π v ) 2
= 14.5 × 9.8 + 14.5 × 1 × ( 2 π × 2 ) 2
= 142.1 + 2 , 289.75 = 2 , 431.85 N
N o w , Y =
orl =
=
= 1.87 × 10 − 3 m = 1.87 m m
When the mass is at the lowest point of its circular path, the stretching force on the wire,
or
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