Calculate moment of inertia w.r.t rotational axis XX’ in following figures
A. $92kg{{m}^{2}},1kg{{m}^{2}}$
B. $32kg{{m}^{2}},1kg{{m}^{2}}$
C. $1kg{{m}^{2}},92kg{{m}^{2}}$
D. $1kg{{m}^{2}},32kg{{m}^{2}}$
Answer
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Hint: In order to answer this question, all you have to have is a basic idea about the moment of inertia. You could thus recall the expression for the moment of inertia of mass m at a certain distance r from the reference axis. You could then substitute the given values and find the answer for both the cases.
Formula used:
Moment of inertia,
$I=M{{R}^{2}}$
Complete answer:
In the question, we are given two figures with some masses and position at which these masses are clearly given. We are supposed to find the moment of inertia about the rotational axis XX’.
Let us recall the very definition of moment of inertia. This quantity is defined as the product of mass and the square of the distance from the mentioned reference axis to the centre of mass. That is, mathematically, this quantity can be given by,
$I=M{{R}^{2}}$
Now for the first figure, we have masses 4kg and 1kg kept at 30cm and 80cm from the given reference axis. As the options are given in SI units let us convert these distances into SI units to get, 0.3m and 0.8m respectively.
So the net moment of inertia of the system will be the sum of the moments of inertia due to 4kg and 1kg masses, that is,
${{I}_{XX'}}=4{{\left( 0.3 \right)}^{2}}+1{{\left( 0.8 \right)}^{2}}$
$\Rightarrow {{I}_{XX'}}=0.36+0.64$
$\therefore {{I}_{XX'}}=1kg{{m}^{2}}$ ……………………………………………….. (1)
Now for the second figure we have three masses and the net moment of inertia of the system about XX’ axis could be given by,
${{I}_{XX'}}=2{{\left( 2 \right)}^{2}}+3{{\left( 4 \right)}^{2}}+4{{\left( 3 \right)}^{2}}$
$\Rightarrow {{I}_{XX'}}=8+48+36$
$\therefore {{I}_{XX'}}=92kg{{m}^{2}}$ …………………………………………………. (2)
From (1) and (2) we found that option C is the correct answer.
Note:
While finding the moment of inertia it is not necessary that you substitute the values in their SI units. But we have to look for the unit given in the options and thus substitute accordingly. Also, distance should be measured from the reference axis about which we calculate the moment of inertia.
Formula used:
Moment of inertia,
$I=M{{R}^{2}}$
Complete answer:
In the question, we are given two figures with some masses and position at which these masses are clearly given. We are supposed to find the moment of inertia about the rotational axis XX’.
Let us recall the very definition of moment of inertia. This quantity is defined as the product of mass and the square of the distance from the mentioned reference axis to the centre of mass. That is, mathematically, this quantity can be given by,
$I=M{{R}^{2}}$
Now for the first figure, we have masses 4kg and 1kg kept at 30cm and 80cm from the given reference axis. As the options are given in SI units let us convert these distances into SI units to get, 0.3m and 0.8m respectively.
So the net moment of inertia of the system will be the sum of the moments of inertia due to 4kg and 1kg masses, that is,
${{I}_{XX'}}=4{{\left( 0.3 \right)}^{2}}+1{{\left( 0.8 \right)}^{2}}$
$\Rightarrow {{I}_{XX'}}=0.36+0.64$
$\therefore {{I}_{XX'}}=1kg{{m}^{2}}$ ……………………………………………….. (1)
Now for the second figure we have three masses and the net moment of inertia of the system about XX’ axis could be given by,
${{I}_{XX'}}=2{{\left( 2 \right)}^{2}}+3{{\left( 4 \right)}^{2}}+4{{\left( 3 \right)}^{2}}$
$\Rightarrow {{I}_{XX'}}=8+48+36$
$\therefore {{I}_{XX'}}=92kg{{m}^{2}}$ …………………………………………………. (2)
From (1) and (2) we found that option C is the correct answer.
Note:
While finding the moment of inertia it is not necessary that you substitute the values in their SI units. But we have to look for the unit given in the options and thus substitute accordingly. Also, distance should be measured from the reference axis about which we calculate the moment of inertia.
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