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Question Number 21038 by ajfour last updated on 10/Sep/17

Commented by ajfour last updated on 10/Sep/17

Find angular velocity and angular  acceleration of rod if end A moves  to the right with constant velocity  v.

$${Find}\:{angular}\:{velocity}\:{and}\:{angular} \\ $$$${acceleration}\:{of}\:{rod}\:{if}\:{end}\:{A}\:{moves} \\ $$$${to}\:{the}\:{right}\:{with}\:{constant}\:{velocity} \\ $$$${v}. \\ $$

Answered by mrW1 last updated on 11/Sep/17

tan θ=(h/x)  (1/(cos^2  θ))×(dθ/dt)=−(h/x^2 )×(dx/dt)=−((h/x))^2 ×(1/h)×(dx/dt)  (1/(cos^2  θ))×(dθ/dt)=−tan^2  θ×(1/h)×(dx/dt)  (dθ/dt)=−sin^2  θ×(1/h)×(dx/dt)  ⇒ angular velocity ω=sin^2  θ×(v/h)  (dω/dt)=2sin θ cos θ (dθ/dt)×(v/h)+sin^2  θ×(dv/dt)  (dv/dt)=0  (dθ/dt)=ω=sin^2  θ×(v/h)  ⇒angular acceleration (dω/dt)=sin 2θ sin^2  θ ((v/h))^2

$$\mathrm{tan}\:\theta=\frac{\mathrm{h}}{\mathrm{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\theta}×\frac{\mathrm{d}\theta}{\mathrm{dt}}=−\frac{\mathrm{h}}{\mathrm{x}^{\mathrm{2}} }×\frac{\mathrm{dx}}{\mathrm{dt}}=−\left(\frac{\mathrm{h}}{\mathrm{x}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{h}}×\frac{\mathrm{dx}}{\mathrm{dt}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\theta}×\frac{\mathrm{d}\theta}{\mathrm{dt}}=−\mathrm{tan}^{\mathrm{2}} \:\theta×\frac{\mathrm{1}}{\mathrm{h}}×\frac{\mathrm{dx}}{\mathrm{dt}} \\ $$$$\frac{\mathrm{d}\theta}{\mathrm{dt}}=−\mathrm{sin}^{\mathrm{2}} \:\theta×\frac{\mathrm{1}}{\mathrm{h}}×\frac{\mathrm{dx}}{\mathrm{dt}} \\ $$$$\Rightarrow\:\mathrm{angular}\:\mathrm{velocity}\:\omega=\mathrm{sin}^{\mathrm{2}} \:\theta×\frac{\mathrm{v}}{\mathrm{h}} \\ $$$$\frac{\mathrm{d}\omega}{\mathrm{dt}}=\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\:\frac{\mathrm{d}\theta}{\mathrm{dt}}×\frac{\mathrm{v}}{\mathrm{h}}+\mathrm{sin}^{\mathrm{2}} \:\theta×\frac{\mathrm{dv}}{\mathrm{dt}} \\ $$$$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{0} \\ $$$$\frac{\mathrm{d}\theta}{\mathrm{dt}}=\omega=\mathrm{sin}^{\mathrm{2}} \:\theta×\frac{\mathrm{v}}{\mathrm{h}} \\ $$$$\Rightarrow\mathrm{angular}\:\mathrm{acceleration}\:\frac{\mathrm{d}\omega}{\mathrm{dt}}=\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\left(\frac{\mathrm{v}}{\mathrm{h}}\right)^{\mathrm{2}} \\ $$

Commented by ajfour last updated on 11/Sep/17

thanks sir,i had missed marking   h in question, thanks for  assuming so and solving perfectly.

$${thanks}\:{sir},{i}\:{had}\:{missed}\:{marking}\: \\ $$$$\boldsymbol{{h}}\:{in}\:{question},\:{thanks}\:{for} \\ $$$${assuming}\:{so}\:{and}\:{solving}\:{perfectly}. \\ $$

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