Wednesday
March 29, 2017

Post a New Question

Posted by on .

A mass 5kg is whirled in a horizontal circle @ 1 end of a string 50cm long,the other end being fixed. If the string when hanging vertically will just support a load of 200kg mass without breaking,find the maximum whirling speed in revolution per second and the maximum angular velocity.

  • physics - ,

    Breaking tensile strength of string = Tmax = 200 kg*9.8 m/s^2 = 1960 N

    Then whirling horizontally,

    T sin A = M g
    T cos A = M V^2/R
    T^2 = M^2*[g^2 + (V^2/R)^2]
    T = Tmax = M* sqrt [g^2 + (V^2/R)^2]
    = 1960 N
    Tmax/M = 392 m/s^2
    = sqrt [g^2 + (V^2/R)^2]
    Solve for V

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question