Question icon
Grade 11Mechanics

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_______________________

Profile image of sindhuja P
16 Years agoGrade 11
Answers icon

1 Answer

Profile image of Badiuddin askIITians.ismu Expert
ApprovedApproved Tutor Answer16 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


Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin