The diameter of a flywheel is increased by $1\%$ . Increase in its moment of inertia about the central axis is:
(A)1%
(B)0.5%
(C)2%
(D)4%
Answer
595.2k+ views
Hint: Firstly we will find the moment of inertia of the flywheel using the formula of moment of inertia of a disk about its central axis. This is given as the product of mass of the flywheel and square of its radius upon two. Then, we shall simplify it and differentiate both sides to get the relative increase in moment of inertia in terms of relative increase in its radius.
Complete answer:
Let the radius of the flywheel be given by (r).
Then, since there is a 1 percent increase in the diameter of the flywheel, it means the radius also increases by 1 percent.
Now, the formula of moment of inertia of a disk or a flywheel about its central axis is given by:
$\Rightarrow I=\dfrac{m{{r}^{2}}}{2}$
Where,
$m$ is the mass of the flywheel
$r$ is the radius of the flywheel
Now, taking logarithms to the base (e) on both sides of the above equation, we get:
$\Rightarrow \ln (I)=\ln \left( \dfrac{1}{2} \right)+\ln (m)+2\ln (r)$
Now, on differentiating both the sides of our equation, we get the new equation as:
$\Rightarrow \dfrac{dI}{I}=0+\dfrac{dm}{m}+2\dfrac{dr}{r}$
Here, the differentiation of the constant $\ln \left( \dfrac{1}{2} \right)$is zero.
Also, since there is no increase or decrease in the mass of the flywheel, it means the mass of the flywheel is constant. Therefore, we can say that:
$\Rightarrow \dfrac{dm}{m}=0$
Using this in the above equation, we get:
$\Rightarrow \dfrac{dI}{I}=2\left( \dfrac{dr}{r} \right)$
Therefore, the percentage change can be written as:
$\Rightarrow \left( \dfrac{dI}{I} \right)\times 100=2\left( \dfrac{dr}{r} \right)\times 100$
We know the percent change in the radius of the flywheel is 1 percent. Hence,
$\Rightarrow \left( \dfrac{dr}{r} \right)\times 100=2$
Putting this value in the above equation, we get:
$\Rightarrow \left( \dfrac{dI}{I}\times 100 \right)=2$
Hence, the percentage increase in the moment of inertia of the flywheel is $2\%$ .
Hence, option (C) is the correct option.
Note:
It should be noted that the moment of inertia of a flywheel should not be confused with the moment of inertia of a ring. This will completely change our answer. These are some basic and commonly used terms in rotation, so one should be aware of their appropriate values.
Complete answer:
Let the radius of the flywheel be given by (r).
Then, since there is a 1 percent increase in the diameter of the flywheel, it means the radius also increases by 1 percent.
Now, the formula of moment of inertia of a disk or a flywheel about its central axis is given by:
$\Rightarrow I=\dfrac{m{{r}^{2}}}{2}$
Where,
$m$ is the mass of the flywheel
$r$ is the radius of the flywheel
Now, taking logarithms to the base (e) on both sides of the above equation, we get:
$\Rightarrow \ln (I)=\ln \left( \dfrac{1}{2} \right)+\ln (m)+2\ln (r)$
Now, on differentiating both the sides of our equation, we get the new equation as:
$\Rightarrow \dfrac{dI}{I}=0+\dfrac{dm}{m}+2\dfrac{dr}{r}$
Here, the differentiation of the constant $\ln \left( \dfrac{1}{2} \right)$is zero.
Also, since there is no increase or decrease in the mass of the flywheel, it means the mass of the flywheel is constant. Therefore, we can say that:
$\Rightarrow \dfrac{dm}{m}=0$
Using this in the above equation, we get:
$\Rightarrow \dfrac{dI}{I}=2\left( \dfrac{dr}{r} \right)$
Therefore, the percentage change can be written as:
$\Rightarrow \left( \dfrac{dI}{I} \right)\times 100=2\left( \dfrac{dr}{r} \right)\times 100$
We know the percent change in the radius of the flywheel is 1 percent. Hence,
$\Rightarrow \left( \dfrac{dr}{r} \right)\times 100=2$
Putting this value in the above equation, we get:
$\Rightarrow \left( \dfrac{dI}{I}\times 100 \right)=2$
Hence, the percentage increase in the moment of inertia of the flywheel is $2\%$ .
Hence, option (C) is the correct option.
Note:
It should be noted that the moment of inertia of a flywheel should not be confused with the moment of inertia of a ring. This will completely change our answer. These are some basic and commonly used terms in rotation, so one should be aware of their appropriate values.
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