Question

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.300 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis: (a) perpendicular to the bar through its center. (b) perpendicular to the bar through one of the balls. (c) parallel to the bar through both balls. (d) parallel to the bar and 0.500 m from it.

Answer 1

Answer:

a)1.93 kg-m^2

b) 1.45  kg-m^2

c) = 0

d) 1.15 kg-m^2

Explanation:

mass of the bar M = 4 kg

length of the bar = 2 m

mass of balls m1= m2= 0.3 kg

moment of inertia of bar $I= \frac{ML^2}{12}$

about an axis perpendicular to the bar through its center.

a) MOI of bar + 2×m×(L/2)^2

$I= \frac{ML^2}{12}$+ $2m\frac{L}{2}^2$

now putting the values of m, M and L as above and solving we get

I= 1.93 kg-m^2

b) perpendicular to the bar through one of the balls

$I=M\frac{L^2}{3} +mL^2$

$I=4\frac{2^2}{3} +0.3\times4^2$= 1.45  kg-m^2

c) parallel to the bar through both balls

zero as the no mass distribution along the parallel to the bar through both balls.

d) parallel to the bar and 0.500 m from it.

I=$(M+2m_1)\frac{1}{2}^2$

putting values and solving we get

1.15 kg-m^2