Question

For a particle executing uniform circular motion with speed v, find the change in velocity during the time interval of (T/6), where T=time period of circular motion.

Answer 1

Dear Student,

$$\begin{gathered} angularfrequency\omega =\frac{2\mathrm{\pi }}{T} \\ letatt=0theparticleisatx=rwhereristheradiusofcircularmotioninX-Yplane. \\ {V}_X=v\sin \omega t \\ {V}_Y=v\cos \omega t \\ velocityatt=0 \\ V=v\hat{j} \\ velocityatt=T/6 \\ V=v\sin \omega t\hat{i}+v\cos \omega t\hat{j}=v\left(\sin \frac{2\mathrm{\pi }}{6}\hat{i}+\cos \frac{2\mathrm{\pi }}{6}\hat{j}\right)=v\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right) \\ changeinvelocity=v\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)-v\hat{j}=v\left(\frac{\sqrt{3}}{2}\hat{i}-\frac{1}{2}\hat{j}\right) \\ Regards \end{gathered}$$