A particle of mass 150 g is attached to one end of a massless inextensible string It is made to describe a vertical circle of radius 1 m When the string is making an angle of `48.2^(@)` with the vertical , its instantaneous speed is ` 2 m//s ` What is the tension in the string in this position ? Whould this particle be able to complete its circular path ? .

Explanation

Here , ` m = 150 g = 0.150 kg ` ,
` r = 1 m , theta = 48.2^(@) , upsilon = 2 m//s , T = ? `
From ` T = (m upsilon^(2))/(r) + mg cos theta`
` T = (0.15 xx 2^(2))/(1) + 0.15 xx 10 xx cos 48 .2^(@) `
` = 0.6 + 1.5 xx 0.6665 = 1.6 N `
To be able to complete the vertical circle ,
` upsilon_(H) = sqrt ( gr) = sqrt(10 xx 1) = 3.16 m//s `
As instantaneous velocity is ` 2 m//s ` it cannot increase (to `3.16 m//s`) on reaching the highest point as gravity is opposing the motion .Therefore , the particle cannot complete its circular path .