20 Dec

9 Comments

  • neema December 20, 2010 at 1:21 pm

    my ans is mgsin(theta)-kR(theta)^2-2mgcos(theta)
    plz let me know if it is correct….

  • neema December 20, 2010 at 1:35 pm

    my first ans is wrong…the ans should be (kR(theta)^2)+(mgsin(theta))

  • neema December 20, 2010 at 1:39 pm

    my ans is kR(theta)^2+(mgsin(theta))
    plz let me know if it is correct….

  • Anudeep December 21, 2010 at 9:34 pm

    isnt the answer so given wrong, how can the tangential force F applied be considered in the contact force

  • rishabh March 2, 2011 at 12:45 am

    sir why don’t you give the SOLUTION..

  • abhishek.hiremath June 22, 2011 at 3:26 pm

    wud somebody mind giving d answers wid d method of solving..:!!

  • Dipak Savaliya September 11, 2011 at 12:38 pm

    it’s answer May be 3mgsin(theta)+kR(theta)^2+2F(theta)

  • lakshya September 11, 2011 at 8:39 pm

    Umm…i dont mind giving the solution…
    The first equation
    1) mgsin theta -N = mv^2/R
    2)The tangential forces on the body will give the equation for its acceleration*mass.
    At an angle theta from the centre :-
    F – (kx+mgcos theta) = m.a
    Here x = R.theta…from angle = arc/radius
    Then…instead of a…we put v.(dv/dx)
    Now it becomes integral of [F – kRtheta – mgcos theta].dx = integral of [m.dv.v]
    replace dx = Rd theta (angle is equal to arc/radius)
    Integrate…get the values of v^2…put them in the first equation…youll get the value of N .:D….really easy 😛

  • sai vishnu September 12, 2011 at 11:23 am

    sir, if the process is given, it would be very helpful for the students.

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