On the marks of 20 cm and 70 cm of a light meter scale, weights of 1kg and 4kg respectively are placed. The moment of inertia in \[kg{{m}^{2}}\] about the vertical axis through the 100 cm mark will be –
A) 1
B) 2
C) 3
D) 4
Answer
601.2k+ views
Hint: The moment of inertia of an object is dependent on its mass at the point under consideration, the axis under consideration and the shape and size of the object. Here, we have a straight rod with an axis along the vertical axis at one end of the rod.
Complete answer:
Moment of inertia is the mass analogous in the rotational motion. It is the resistance offered by the mass against the rotation along a given axis. The center of mass of the object under consideration needs to be determined to calculate the moment of inertia. The center of mass of a system of masses is given by –
\[\begin{align}
& \text{Center of mass,} \\
& CM=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{r}_{i}}}}{M} \\
\end{align}\]
Where m is the mass of individual points,
r is the distance from the origin.
M is the total mass of the system.
We can find the moment of inertia of a system as –
\[\begin{align}
& \text{Moment of inertia,} \\
& I=\sum\limits_{i=1}^{n}{{{m}_{i}}{{r}_{i}}^{2}} \\
\end{align}\]
Now, we know that the moment of inertia depends on the position of the axis. For our given situation it is at the one end along the vertical. The moment of inertia for such an arrangement is the same as the basic formula.
i.e.,
\[\begin{align}
& \text{Moment of inertia,} \\
& I=\sum\limits_{i=1}^{n}{{{m}_{i}}{{r}_{i}}^{2}} \\
\end{align}\]
Now, we are given two points at two marks on a scale. They are –
\[\begin{align}
& {{m}_{1}}=1kg,{{r}_{1}}=0.8m \\
& {{m}_{2}}=4kg,{{r}_{2}}=0.3m \\
\end{align}\]
Now, let us substitute this in the formula to find the moment of inertia –
\[\begin{align}
& I={{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2} \\
& \Rightarrow \text{ }I=1\times {{(0.8)}^{2}}+4\times {{(0.3)}^{2}} \\
& \Rightarrow \text{ }I=0.64+0.36 \\
& \therefore \text{ }I=1kg{{m}^{-2}} \\
\end{align}\]
The moment of inertia is 1\[kg{{m}^{-2}}\]
So, the correct answer is “Option A”.
Additional Information:
The moment of inertia varies with the position of center of mass.
Note:
The moment of inertia is calculated based on the distance from its center of mass and the mass involved. According to the shape, size and axis selected, the length term in the moment of inertia keeps on changing. For a rod of similar features as we have seen now, if the axis is taken perpendicular, then the moment of inertia reduces to one-twelfth of earlier.
Complete answer:
Moment of inertia is the mass analogous in the rotational motion. It is the resistance offered by the mass against the rotation along a given axis. The center of mass of the object under consideration needs to be determined to calculate the moment of inertia. The center of mass of a system of masses is given by –
\[\begin{align}
& \text{Center of mass,} \\
& CM=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{r}_{i}}}}{M} \\
\end{align}\]
Where m is the mass of individual points,
r is the distance from the origin.
M is the total mass of the system.
We can find the moment of inertia of a system as –
\[\begin{align}
& \text{Moment of inertia,} \\
& I=\sum\limits_{i=1}^{n}{{{m}_{i}}{{r}_{i}}^{2}} \\
\end{align}\]
Now, we know that the moment of inertia depends on the position of the axis. For our given situation it is at the one end along the vertical. The moment of inertia for such an arrangement is the same as the basic formula.
i.e.,
\[\begin{align}
& \text{Moment of inertia,} \\
& I=\sum\limits_{i=1}^{n}{{{m}_{i}}{{r}_{i}}^{2}} \\
\end{align}\]
Now, we are given two points at two marks on a scale. They are –
\[\begin{align}
& {{m}_{1}}=1kg,{{r}_{1}}=0.8m \\
& {{m}_{2}}=4kg,{{r}_{2}}=0.3m \\
\end{align}\]
Now, let us substitute this in the formula to find the moment of inertia –
\[\begin{align}
& I={{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2} \\
& \Rightarrow \text{ }I=1\times {{(0.8)}^{2}}+4\times {{(0.3)}^{2}} \\
& \Rightarrow \text{ }I=0.64+0.36 \\
& \therefore \text{ }I=1kg{{m}^{-2}} \\
\end{align}\]
The moment of inertia is 1\[kg{{m}^{-2}}\]
So, the correct answer is “Option A”.
Additional Information:
The moment of inertia varies with the position of center of mass.
Note:
The moment of inertia is calculated based on the distance from its center of mass and the mass involved. According to the shape, size and axis selected, the length term in the moment of inertia keeps on changing. For a rod of similar features as we have seen now, if the axis is taken perpendicular, then the moment of inertia reduces to one-twelfth of earlier.
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