Sum

A wheel of moment of inertia 0⋅500 kg-m^{2} and radius 20⋅0 cm is rotating about its axis at an angular speed of 20⋅0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.

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#### Solution

Given

Initial moment of inertia of the system,

I_{1} = 0.500 kg-m^{2};

r = 0.2 m;

ω = 20 rad/s

Mass of the stationary particle, m = 0.2 kg

Final moment of inertia of the system,

I_{2} = I_{1}_{ }+ mr^{2}

It is given

External torque = 0

Angular momentum is conserved; therefore, we have

\[I_1 \omega_1 = I_2 \omega_2\]

\[\Rightarrow 0 . 5 \times 20 = \left( 0 . 5 + 0 . 2 \times \left( 0 . 2 \right)^2 \right) \omega_2 \]

\[ \Rightarrow \omega_2 = \frac{10}{0 . 508} \approx 19 . 7\text{ rad/s}\]

Concept: Moment of Inertia

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