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Standard XII

Physics

Angular Velocity

A turntable h...

Question

A turntable has a radius of $0.80 m$ and a moment of inertia of $2.0 k g m^{2}$ . The turntable is rotating with an angular velocity of $1.5 r a d / s$ about a vertical axis through its center on frictionless bearings. A very small $0.40 - k g$ ball is projected horizontally towards the turntable axis with a velocity of $3.0 m / s$ . The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure). What is the angular velocity of the turntable just after the ball is caught ?

2.1 r a d / s

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1.3 r a d / s

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0.94 r a d / s

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0.75 r a d / s

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Solution

The correct option is B $1.3 r a d / s$
$\begin{matrix} Geven I = 2 {k g m}^{2} R = 0.8 m ω_{0} = 1.5 r a d / m = 0.4 k g Using conservation of angular momentum \end{matrix}$
$\begin{matrix} L_{i} = L_{f} I ω_{0} = I_{1} ω_{f} + L_{b c l l} 2 (1.5) = (2 + ω + m r^{2} ω 3 = 2 ω + (0.4) (0.8)^{2} ω ω = \frac{3}{2.256} ω = 1.31 r a d / s \end{matrix}$

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A man of mass 100 kg stands at the rim of a turntable of radius 2m and moment of inertia 4000 $k g - m^{2}$ mounted on a vertical frictionless shaft at its center. The whole system is initially at rest. The man now walks along the outer edge of a turntable with a velocity of 1 m/s relative to the earth. With what angular velocity and in what direction does the turntable rotates?

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The correct option is B 1.3rad/s Geven I=2kgm2R=0.8 mω0=1.5rad/m=0.4 kg Using conservation of angular momentum Li=LfIω0=I1ωf+Lbcll2(1.5)=(2+ω+mr2ω3=2ω+(0.4)(0.8)2ωω=32.256ω=1.31rad/s