an object of mass 0.5 is whirled at the end of a string 0.8 m long.if the string makes three revolution in 1.2 seconds, find the tension on

Asked by devdrondiksha | 24th May, 2022, 03:27: PM

Expert Answer:

Given that,
Mass of object, m = 0.5 kg
length of string, l = 0.8 m 
Number of revolution in 1.2 sec, n= 3
Angular velocity, omega space equals fraction numerator 2 πn over denominator t end fraction equals fraction numerator 2 straight pi cross times 3 over denominator 1.2 end fraction equals space 5 space r a d divided by s e c
Now,
Tension on string will be 
T space almost equal to fraction numerator m v squared over denominator r end fraction equals m omega squared r
i. e. comma space T equals space 0.5 space cross times left parenthesis 5 straight pi right parenthesis squared cross times 0.8
rightwards double arrow T space almost equal to space 158 space N

Answered by Jayesh Sah | 26th May, 2022, 11:32: AM

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