Question #160706

Three particles at the vertices of an equilateral triangle move towards each other....find the initial acceleration of the particles


1
Expert's answer
2021-02-02T15:34:17-0500

Three particles at the vertices of an equilateral triangle move towards each other.

velocity magnitude of the particles remains the same throughout the motion ,velocity of each particle would have its component VCos30°V Cos 30° towards the center and VSin30°V Sin30° perpendicular to it. Distance of the centre from its vertex at the beginning is SS Cos30°×23=S3Cos30°×\large\frac{2}{3}=\sqrt{\frac{S}{3}} The component VSin30°=V2VSin30°=\large\frac{V}{2} is forms tangential component on the circular path described by each particle. Thus the initial centripetal acceleration is

a=(VSin30°)2r=VS3.(Since r=S3)=(V2)×34Sa=\large\frac{{}{}(V Sin 30°)^2}{r}=\frac{\sqrt{V}}{\sqrt{\frac{S}{3}}} …………. (Since \space r = \sqrt{\frac{S}{3}}) =(V^2)×\frac{\sqrt{3}}{4S}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS