A flywheel of mass $1 \mathrm{~kg}$ and radius vector $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2…

A flywheel of mass $1 \mathrm{~kg}$ and radius vector $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mathrm{m}$ is at rest. When a force $(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \mathrm{N}$ acts on it tangentially, it can rotate freely. Then, its angular velocity after $4.5 \mathrm{~s}$ is
  1. $\frac{2}{9} \sqrt{261} \mathrm{rad} \mathrm{s}^{-1}$
  2. $\frac{3}{2} \sqrt{261} \mathrm{rad} \mathrm{s}^{-1}$
  3. $\sqrt{261} \mathrm{rad} \mathrm{s}^{-1}$
  4. $\frac{5}{9} \sqrt{261} \mathrm{rad} \mathrm{s}^{-1}$

Solution

Given, mass of flywheel, $M=1 \mathrm{~kg}$, radius vectors $\mathbf{R}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mathrm{m}$, $\mathbf{F}=(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \mathrm{N}$ and time, $t=4.5 \mathrm{~s}$ Magnitude of radius $ (R)=\sqrt{(2)^2+(1)^2+(2)^2}=\sqrt{9}=3 \mathrm{~m} $ Similarly, $F=\sqrt{29} \mathrm{~N}$ Torque on the flywheel, $ \begin{gathered} \tau=I \alpha=F \cdot R=\frac{M R^2}{2} \alpha \\ \alpha=\frac{2 F}{M R}=\frac{2 \sqrt{29}}{1 \times 3}=\frac{2}{3} \sqrt{29} \end{gathered} $ Now, the angular velocity, $ \begin{aligned} \omega & =\omega_0+\alpha t \Rightarrow \omega=0+\frac{2}{3} \sqrt{29} \times 4.5\left(\because \omega_0=0\right) \\ \Rightarrow \quad \omega & =\sqrt{261} \mathrm{rad} \mathrm{s}^{-1} \end{aligned} $ Hence, the correct option is (c)

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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