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A particle of mass 2Kg revovles in a circular path on X-Y plane with a constant speed of 5 m/s along the path x 2 +y 2 -25=0. The force(in N) acting on the particle at the position (-3m,4m) will be_______________________

A particle of mass 2Kg revovles in a circular path on X-Y plane with a constant speed of 5 m/s along the path x2+y2-25=0.The force(in N) acting on the particle at the position (-3m,4m) will be_______________________

Grade:11

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear sindhuja

 

6386-1362_7662_untitled1.JPG

 

radius of the circle is =5

since it  is moving with constant speed so there is no tangential acceleration

only centripetal acceleraton will act towards the center of circle .

centripetal force =mv2/r

                       =2*25/5 =10 N

tan θ =|4/-3|

        =4/3

so force =10(cosθ i -sinθj)

            =6i-8j


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Badiuddin

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