Question

a string of length 1m is fixed at one end and carries mass of 100gm at the other end . The string makes 2÷Πrevolutions per second around the vertical axis through the fixed end . calculate the tenstion in the string​

Answer 1

Answer:

1.6 N

Explanation:

Tcos∝ = mg

Tsin∝ = mv²/R

tan∝ = v²/Rg = R ÷ √1 - R² = w²R/g , √1 - R² = g/w²

w = 2π/T = 2πf = 2π × 2/π = 4 rad/s

√1 - R² = 10/16 = 5/8 , R² = 1 - 25/64 = 39/64 , R ≈ 0.8 m

tan∝ = 1.28 , ∝ = 52°

T = mg/cos∝ = 0.1×10/cos 52° = sec 52² = 1.6N

Answer 2

Answer:

1.6N

Explanation:

Tsin thetha =m*v*v/r -----------------(1)

Tcos thetha = m*g --—----------------(2)

W=2 /Pi * 2pi = 4 radis

From 1 and 2

Tan thetha = v^2/rg

= w^2r/g

Sin thetha = r

Tan thetha = 4^2*(sin thetha) /g

Sin thetha/ cos thetha =16*sin thetha /10

1/cos thetha = 1/1.6 -----------------(3)

= 0.625

Thetha= 52°

From ( 3)

T= mg/cos thetha

=(0.1*10) /(1/1.6)

T= 1.6 N