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Question Number 80341 by Rio Michael last updated on 02/Feb/20

$$\mathrm{A}\mathrm{particle}\mathrm{moves}\mathrm{round}\mathrm{the}\mathrm{polar}\mathrm{curve}$$$$r=a(1+\cos \theta )\mathrm{with}\mathrm{constant}\mathrm{angular}$$$$\mathrm{velocity}\omega .\mathrm{Find}\mathrm{the}\mathrm{transverse}\mathrm{component}$$$$\mathrm{of}\mathrm{the}\mathrm{velocity}.$$

Answered by mr W last updated on 02/Feb/20

Commented by mr W last updated on 02/Feb/20

$$r=a(1+\cos \theta )$$$$\omega =\frac{d\theta }{dt}=constant$$$$v_T=v_{\theta }=\frac{rd\theta }{dt}=\omega a(1+\cos \theta )=tranversevelocity$$

Commented by Rio Michael last updated on 02/Feb/20

$$thankssir$$

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