Question

A string of length 10m is fixed at one end and carries a mass of 100g and other end the string makes 2 by pii revolution /sec around vertical axis passes through its second end, Find - 1)Angle of inclination of string with the vertical - 2)Tension in the string -3)Velocity of mass

Answer 1

The angle of inclination is θ = cos^-1 (1/10), Tension in the string is T = 16 N and velocity of mass is V = 40 m/s.

Explanation:

mω^2.R =  T.sinθ

mω^2 x 10 sinθ = T.sinθ

T = 10 mω^2

ω = 2 / π x 2π rad / sec

ω = 4 rad / sec

10 mω^2 = 10 x 100 / 1000 x (4)^2

T = 16 N

T cos θ = mg

cos θ = mg / T = 100 / 1000 x 10 ÷ 16 = 1/16

θ = cos^-1 (1/10)

V^2 = ω^2.R^2

V^2 = (4)^2 (10)^2

V = √ (4)^2 (10)^2  

V = 40 m/s

Thus the angle of inclination is θ = cos^-1 (1/10), Tension in the string is T = 16 N and velocity of mass is V = 40 m/s.

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