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A conical pen...

Question

A conical pendulum of length $1 m$ makes an angle $θ = 45^{o}$ w.r.t. Z-axis and moves in a circle in the $X Y$ plane. The radius of the circle is $0.4 m$ and its center is vertically below $O$ . The speed of the pendulum, in its circular path, will be : (Take $g = 10 m s^{- 2}$ )

0.4 m / s

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4 m / s

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0.2 m / s

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2 m / s

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Solution

The correct option is A $0.4 m / s$
$l = 1 m$ $r = 0.4 m$ $θ = 45^{\circ}$
$F = \frac{m v^{2}}{r} [\frac{T}{T_{2}}] = m g [\frac{T}{T_{2}}] = F F = m g \frac{m v^{2}}{r} = m g v^{2} = r g = 0.4 \times 10 = 4 v = 2 m / s$

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A conical pendulum of length 1 m makes an angle θ=45o w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be : (Take g=10ms−2)

The correct option is A 0.4 m/sl=1m r=0.4m θ=45∘ F=mv2r[TT2]=mg[TT2]=FF=mgmv2r=mgv2=rg=0.4×10=4v=2m/s

A conical pendulum of length $1 m$ makes an angle $θ = 45^{o}$ w.r.t. Z-axis and moves in a circle in the $X Y$ plane. The radius of the circle is $0.4 m$ and its center is vertically below $O$ . The speed of the pendulum, in its circular path, will be : (Take $g = 10 m s^{- 2}$ )

The correct option is A $0.4 m / s$
$l = 1 m$ $r = 0.4 m$ $θ = 45^{\circ}$
$F = \frac{m v^{2}}{r} [\frac{T}{T_{2}}] = m g [\frac{T}{T_{2}}] = F F = m g \frac{m v^{2}}{r} = m g v^{2} = r g = 0.4 \times 10 = 4 v = 2 m / s$