Question

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is 60.0 cm long and has mass 0.400 kg.

(a) What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod?
(b) One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a 60.0∘ angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the V at its vertex?

Answer 1

Answer:

a) $I=0.012\ kg.m^2$

b) $I=0.012\ kg.m^2$

Explanation:

Given that:

• mass of rod, $m=0.4\ kg$

• length of the rod, $l=0.6\ kg$

(a)

Moment of inertia of rod about its center and perpendicular to the rod is given as:

$I=\frac{1}{12} m.l^2$

$I=\frac{1}{12} 0.4\times 0.6^2$

$I=0.012\ kg.m^2$

(b)

Moment of inertia on bending the rod to V-shape of 60 degree angle and axis being perpendicular to the plane of V at the vertex.

We treat it as two rod with axis of rotation at the end and perpendicular to the plane of rotation.

So, the mass and the length of the rod will become half of initial value.

$I=I_1+I_2$

$I=\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2 +\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2$

$I=2[ \frac{1}{3}\times 0.2\times 0.3^2]$

$I=0.012\ kg.m^2$

Answer 2

The moment of inertia of this rod for an axis at its center and the of the bent rod about an axis perpendicular to the plane of the V at its vertex is mathematically given as

• I=0.012 kg.m^2
• I'=0.012 kg.m^2

What is the moment of inertia of this rod for an axis at its center and the of the bent rod about an axis perpendicular to the plane of the V at its vertex?

Question Parameter(s):

a uniform rod that is 60.0 cm long and has mass 0.400 kg.

Generally, the equation for the Moment of inertia  is mathematically given as

$I=\frac{1}{12} m.l^2$

These

I=1/12 0.4* 0.6^2

I=0.012kg.m^2

In conclusion, the bent rod

I=I_1+I_2

I=1/3*m/2*(l/2)^2 +(1/3) (m/2) (0.5)^2

I=2[ 1/3* 0.2* 0.3^2]

I'=0.012 kg.m^2

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