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A Bob of simple pendulum is given horizontal velocity of √10gl when it is at lowest position. Find the tension in the string when string rotates by 120°

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6 months ago

Arun
24737 Points
```							Now assume that the position of bob is upward horizontal.(a)  Let the velocity at x = v₂ Now K.E at v₁ = 1/2 mv₁² K.E at v₂ = 1/2mv₂²P.E = mgl Now kinetic energy at this point is equal to total energy, Equate both1/2 mv₁² = 1/2 mv₂² + mgl -----→(i)1/2 m√(10)² = 1/2 mv₂² + mglv₂² = 8 gl .Now the tension at the horizontal position. (b) Assume that the velocity at y = v₃Similarly like equation (i) , at y and velocity v₃1/2mv₁² = 1/2mv₃² + mg(2 l ) v₃² = 6mglNow thew tension -mg × mg (c)Assume the velocity at z = v₄ .Again similarly like equation (i) at z = v₄1/2mv₁² = 1/2mv₄² +mgh    ∵[h = l(1+cosθ)]Now, 1/2 mv₁² =1/2 mv₄² +mgl(1+cos60°)calculating it , we getv₄² =7glNow the tension , -mg(cos60°) .On calculating , we get = 7mg -0.5mg  = 6.5 mg
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6 months ago
Vikas TU
12121 Points
```							Dear student The above ans is missing the formulas and steps Here is the complete solution Let tension = T Mass ( bob) = m  velocity = v.Now assume that the position of bob is upward horizontal.(a)  Let the velocity at x = v₂ Now K.E at v₁ = 1/2 mv₁² K.E at v₂ = 1/2mv₂²P.E = mgl Now kinetic energy at this point is equal to total energy, Equate both1/2 mv₁² = 1/2 mv₂² + mgl -----→(i)1/2 m√(10)² = 1/2 mv₂² + mglv₂² = 8 gl .Now the tension at the horizontal position.T = mv^2 /r T = m8gl/l  (b) Assume that the velocity at y = v₃Similarly like equation (i) , at y and velocity v₃1/2mv₁² = 1/2mv₃² + mg(2 l )1/2 m(logl) = 1/2 mv3^2 + 2mgl  v₃² = 6mglNow thew tension Ty = mv^2 /l -mg T = 6mgl/l × mgT = 5mg (c)Assume the velocity at z = v₄ .Again similarly like equation (i) at z = v₄1/2mv₁² = 1/2mv₄² +mgh    ∵[h = l(1+cosθ)]Now,1/2 mv₁² =1/2 mv₄² +mgl(1+cos60°)calculating it , we getv₄² =7glNow the tension , T2 = mv^2/l -mg(cos60°) .On calculating , we get T2 = 7mg -0.5mg  = 6.5 mg
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6 months ago
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