Question

Three identical particles are fixed to the corners of an isosceles right-angled triangle by the means of massless connecting rods. Each of the two sides has a length d, the moment of inertia of this rigid object when the axis of rotation coincides with the hypotenuse of the triangle is..
a] 1/2 md^2
b]1/4md^2
c]md^2
d}3/4md^2

Answer 1

Given:

Three identical particles are fixed to the corners of an isosceles right-angled triangle by the means of massless connecting rods. Each of the two sides has a length d.

To find:

Moment Of Inertia when the axis is along the hypotenuse.

Calculation:

Since the axis is along the hypotenuse only the particle at the vertex opposite to the hypotenuse will contribute to moment of inertia.

Let perpendicular distance of that particle from the axis be x ;

$\therefore \: \dfrac{1}{2} \times d \times d = \dfrac{1}{2} \times x \times ( \sqrt{ {d}^{2} + {d}^{2} } )$

$= > \: \dfrac{1}{2} \times {d}^{2} = \dfrac{1}{2} \times x \times ( \sqrt{ 2 {d}^{2} } )$

$= > \: {d}^{2} = x \times ( \sqrt{ 2 {d}^{2} } )$

$= > \: {d}^{2} = x \times d\sqrt{2}$

$= > \: x = \dfrac{d}{ \sqrt{2} }$

Now , required moment of inertia:

$\therefore \: MI = m {x}^{2}$

$= > \: MI = m {( \dfrac{d}{ \sqrt{2} }) }^{2}$

$= > \: MI = m \dfrac{ {d}^{2} }{2}$

$= > \: MI = \dfrac{1}{2} m {d}^{2}$

So, final answer is:

$\boxed{ \bf{\: MI = \dfrac{1}{2} m {d}^{2} }}$