Question

Three identical discs, each having mass 'M' and diameter 'd', are touching each other as shown in the figure. Calculate the ratio of moment of inertia of the system of three discs about an axis perpendicular to the plane of the paper and passing through point P and B as shown in the figure. Given P is the centroid of the triangle. (See attachment)
A. 11/20
B. 11/18
C. 11/17
D. 11/19

Answer 1

To find:

Moment of inertia when axis passes through point P ?

Calculation:

First of all, we need to find the value of AP, BP, CP:

• We know that centroid divides altitude in the ratio of 2 : 1.

• Also, height of equilateral ∆ is (√3)s/2.

• In the figure, equilateral ∆ has side = d.

$\rm AP = BP = CP = \dfrac{2}{3} \times \dfrac{ \sqrt{3}d }{2}$

$\rm \implies AP = BP = CP = \dfrac{d}{ \sqrt{3} }$

Now, we need to apply PARALLEL AXIS THEOREM for all the 3 discs:

$\rm MI = 3 \times \bigg \{ \dfrac{M {( \frac{d}{2}) }^{2} }{2} + M { (\dfrac{d}{ \sqrt{3} }) }^{2} \bigg \}$

$\rm \implies MI = 3 \times \bigg \{ \dfrac{M {d}^{2} }{8} + \dfrac{M {d}^{2} }{3} \bigg \}$

$\rm \implies MI = 3 \times \bigg \{ \dfrac{(3 + 8)M {d}^{2} }{24} \bigg \}$

$\rm \implies MI = 3 \times \bigg \{ \dfrac{11M {d}^{2} }{24} \bigg \}$

$\rm \implies MI = \dfrac{11M {d}^{2} }{8}$

So, final answer is:

$\boxed{ \bf MI = \dfrac{11M {d}^{2} }{8}}$