Grade 12th passMechanics

a particle of mass 100gm is made to describe a vertical circle of 1m. its instantaneous speed is 1m/s when the string makes an angle of 30 degree with the vertical. find the tension in the string at this position .can the particle complete its circular path ?

Shaima Nishar

9 Years agoGrade 12th pass

1 Answer

Piyush Kumar Behera

9 Years ago

mass = 0.1 kg

drawing the fbd at that instant,we get the following equation,

T-mgcos30=mv²/r

because this only provides the centripetal force.

and here v is the instantaneous speed of the particle =1m/s and r=1

So T=mv²/r+mgcos30=0.1N+0.866N=0.966N (substituting the values)

to complete the loop it must have a total mechanical energy of 2.5mgr (reference for potential enrgy is the lowest point in the circle,the minimum velocity is $\sqrt{5gr}$ )

total mechanical enrgy at that point is m/2+1.34m=0.84m (taking r=1 and m the mass)

so it cannot complete the circle.

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a particle of mass 100gm is made to describe a vertical circle of 1m. its instantaneous speed is 1m/s when the string makes an angle of 30 degree with the vertical. find the tension in the string at this position .can the particle complete its circular path ?

Shaima Nishar

1 Answer

Piyush Kumar Behera

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