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 \omega _0 is minimum, is given by :

  • Option 1)

    x = \frac{{\text{m}_1 }} {{\text{m}_2 }}\text{L}

  • Option 2)

    x = \frac{{\text{m}_2 }} {{\text{m}_1 }}\text{L}

  • Option 3)

    x = \frac{{\text{m}_2 \text{L}}} {{\text{m}_1 + \text{m}_2 }}

  • Option 4)

    x = \frac{{\text{m}_1 \text{L}}} {{\text{m}_1 + \text{m}_2 }}

 

Answers (1)

As we learned in concept

Moment of inertia for system of particle -

I= m_{1}r_{1}^{2}+m_{2}r_{2}^{2}+.........m_{n}r_{n}^{2}

dpi{100} = sum_{i=1}^{n}: m_{i}r_{i}^{2}

 

- wherein

Applied when masses are placed discretely.

 

 Total M.I of rod

I = m_{1}x^{2}+m_{2}(L-x)^{2}

I = m_{1}x^{2}+m_{2}L^{2}+m_{2}x^{2}-m_{2}Lx

As I is minimum i.e.

\frac{di}{dx} = 2m_{1}x+0+2xm_{2}-2m_{2}L=0

=>x(2m_{1}+2m_{2})= 2m_{2}L

x = \frac{m_{2}L}{m_{1}+m_{2}}


Option 1)

x = \frac{{\text{m}_1 }} {{\text{m}_2 }}\text{L}

this is the incorrect option

Option 2)

x = \frac{{\text{m}_2 }} {{\text{m}_1 }}\text{L}

this is the incorrect option

Option 3)

x = \frac{{\text{m}_2 \text{L}}} {{\text{m}_1 + \text{m}_2 }}

this is the correct option

Option 4)

x = \frac{{\text{m}_1 \text{L}}} {{\text{m}_1 + \text{m}_2 }}

this is the incorrect option

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