Point masses m₁ and m₂ are placed at the opposite ends of a rigid rod of length L,

Point masses m₁ and m₂ are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω₀ is minimum, is given by

Options

(a) x= m₂/m₁ L
(b) x= m₂L/m₁+m₂
(c) x= m₁L/m₁+m₂
(d) x= m₁/m₂ L

Correct Answer:

x= m₂L/m₁+m₂

Explanation:

No explanation available. Be the first to write the explanation for this question by commenting below.

admin:

Related Questions

  1. A paramagnetic sample shows a net magnetisation of 0.8 Am⁻¹, when placed
  2. Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line
  3. A car accelerates from rest at constant rate for first 10s and covers a distance x.
  4. For a given velocity, a projectile has the same range R for two angles of projection.
  5. The magnetic moment of a diamagnetic atom is