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Question Number 140738 by BHOOPENDRA last updated on 12/May/21

Commented by BHOOPENDRA last updated on 12/May/21

$$mr.Wsir?$$

Commented by mr W last updated on 13/May/21

$$\delta \theta =\omega \delta t$$$$\delta p=r\delta \theta =\frac{b}{2}\delta \theta =\frac{b\omega \delta t}{2}$$$$\frac{\delta y}{\delta p}=\frac{\frac{\delta p}{2}}{a}=\frac{\delta p}{2a}$$$$\Rightarrow \delta y=\frac{{\left(\delta p\right)}^2}{2a}=\frac{b^2{\left(\omega \delta t\right)}^2}{8a}\Rightarrow \left(A\right)iscorrect$$$$$$$$accelerationinverticaldirection=A$$$$sincey=\frac{1}{2}{At}^2$$$$\Rightarrow \delta y=\frac{1}{2}A{\left(\delta t\right)}^2$$$$A=\frac{2\delta y}{{\left(\delta t\right)}^2}=\frac{b^2{\omega }^2}{4a}$$$$(F+dF)-F=mA$$$$\Rightarrow dF=mA=\frac{{mb}^2{\omega }^2}{4a}\Rightarrow \left(C\right)iscorrect$$

Commented by BHOOPENDRA last updated on 13/May/21

$$optionaandcwillbecorrect$$$$andcanyouexplainhowyoutheseequation?$$

Commented by mr W last updated on 13/May/21

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