NEET-XII-Physics
07: Circular Motion
- #15Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use cos θ ≈ 1 - θ2/2 and SINθ ≈ θ for small θ.Ans : Given:
Mass of the bob = m = 0.1 kg
Length of the circle = R = 1 m
Velocity of the bob = v = 1.4 m/s
Let T be the tension in the string when it makes an angle of 0.20 radian with the vertical.
From the free body diagram, we get:
`` T-mg\,\mathrm{\,cos \,}\theta =\frac{m{v}^{2}}{R}``
`` T=\frac{m{v}^{2}}{R}+mg\,\mathrm{\,cos \,}\theta ``
`` \,\mathrm{\,For \,}\,\mathrm{\,small \,}\,\mathrm{\,\theta \,},\,\mathrm{\,it \,}\,\mathrm{\,is \,}\,\mathrm{\,given \,}\,\mathrm{\,that \,}:``
`` \,\mathrm{\,cos \,}\theta =1-\frac{{\theta }^{2}}{2}``
`` \therefore T=\frac{0.1\times (1.4{)}^{2}}{1}+(0.1)\times 9.8\left(1-\frac{{\theta }^{2}}{2}\right)``
`` =0.196+0.98\times \left(1-\frac{{\left(0.2\right)}^{2}}{2}\right)``
`` =0.196+0.9604``
`` =1.156\,\mathrm{\,N \,}\approx 1.16\,\mathrm{\,N \,}``
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