The rigid body shown in the fig- a has a mass of 10 kg- It rotates with a uniform angular velocity - A balancing mass of 20 kg is attached as shown in fig-b- The percentage increase in mass moment of inertia as a result of this addition is

Moment of inertia of first nbsp;=mk^2=10× (0.2)^2 =0.4 kg.m^2 Moment of inertia of second =m_1k_1^2+m_2k_2^2 =10 ×(0.2)^2 +20×(0.1)^2 =0.6 kg.m^2 ∴Increases ...

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Engineering Mechanics

Mass Moment of Inertia of Standard Objects

The rigid bod...

Question

The rigid body shown in the fig. (a) has a mass of 10 kg. It rotates with a uniform angular velocity $ω$ . A balancing mass of 20 kg is attached as shown in fig.(b). The percentage increase in mass moment of inertia as a result of this addition is

25%

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50%

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100%

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200%

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Solution

The correct option is B
50%

Moment of inertia of first
$= m k^{2} = 10 \times (0.2)^{2}$
$= 0.4 k g . m^{2}$

Moment of inertia of second
$= m_{1} k_{1}^{2} + m_{2} k_{2}^{2}$
$= 10 \times (0.2)^{2} + 20 \times (0.1)^{2}$
$= 0.6 k g . m^{2}$

$∴ Increases in mass moment of inertia$
$= \frac{0.6 - 0.4}{0.4} \times 100 = 50$ %

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The rigid body shown in the fig. (a) has a mass of 10 kg. It rotates with a uniform angular velocity ω. A balancing mass of 20 kg is attached as shown in fig.(b). The percentage increase in mass moment of inertia as a result of this addition is

The correct option is B 50% Moment of inertia of first =mk2=10×(0.2)2 =0.4kg.m2 Moment of inertia of second =m1k21+m2k22 =10×(0.2)2+20×(0.1)2 =0.6kg.m2 ∴Increases in mass moment of inertia =0.6−0.40.4×100=50%

The rigid body shown in the fig. (a) has a mass of 10 kg. It rotates with a uniform angular velocity $ω$ . A balancing mass of 20 kg is attached as shown in fig.(b). The percentage increase in mass moment of inertia as a result of this addition is