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

Answer 2

The moment of inertia of the rod ball system as the mass and distance

from axis of rotation increases.

The correct responses for the moment of inertia are;

(a) Perpendicular the bar and through the bar's center; $\underline{1.9\overline 3 \ kg\cdot m^2}$

(b) Perpendicular the bar and through the ball $\underline{10.1\overline 3 \ kg \cdot m^2}$

(c) 0

(d) Parallel to the bar 1.15 kg·m²

Reasons:

(a) The moment of inertia of the bar perpendicular to the bar through its

center is given by, the equation;

$I_{\perp \odot } = I_{bar} + I_{ball \ on \ left} + I_{ball \ on \, right} = \dfrac{1}{12} \cdot m_{bar}\cdot L^2 + m \cdot r^2 + m \cdot r^2$

Therefore;

$I_{\perp \odot }= \dfrac{1}{12} \times 4\times 2^2 + 0.3 \times 1^2 + 0.3 \times 1^2 = \dfrac{29}{15} \approx 1.9\overline 3$

$I_{\perp \odot } \approx \underline{1.9\overline 3 \ kg\cdot m^2}$

(b) The moment of inertia perpendicular to the bar through one of the balls is perpendicular to the bar through one of the ends is given as follows;

$I_{\perp end } = \dfrac{1}{3} \cdot m_{bar} \cdot L^2 + m\cdot L^2 + m\cdot 0^2$

$I_{\perp end } = \dfrac{1}{3} \times 4\times 2^2 + 0.3\times 4^2 + m\cdot 0^2 = 10.1\overline 3$

$I_{end \perp } = \underline{10.1\overline 3 \ kg \cdot m^2}$

(c) The moment of inertia through the balls and parallel to the bar is the moment of inertia through the axis of the bar, given as follows;

$I_{CM} = \dfrac{1}{2} \cdot m_{bar} \cdot r_{bar}^2 + 2 \times \dfrac{2}{5} \cdot m_{ball} \cdot r_{ball}^2$

$r_{bar}$ = 0

$r_{ball}$ = 0

Therefore;

$I_{CM} = \dfrac{1}{2} \cdot m_{bar} \times 0^2 + 2 \times \dfrac{2}{5} \cdot m_{ball} \times 0^2 = 0 + 0$

The moment of inertia through both balls and parallel to the bar = 0

(d)  By parallel axis theorem, we have;

$I_{CM + d}$ = $I_{CM}$ + $m_{bar}$·d² + $m_{ball}$·d² + $m_{ball}$·d²

∴  $I_{CM + d}$ = 0 + 4×0.5² + 0.3×0.5² + 0.3×0.5² = 1.15

$I_{CM + d}$ = 1.15 kg·m²

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