
Let the moment of inertia of a hollow cylinder if length 10 cm ( inner radius 10 cm and outer radius 20 cm) about its axis be I . The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I is ____
A. 12cm
B. 18cm.
C. 16cm
D. 14cm
Answer
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Hint: Apply the formula for the moment of inertia of the hollow cylinder with inner and outer radius and find out the value of the moment of inertia of the hollow cylinder and then find out the moment of inertia of a thin cylinder with some assumed radius and equate the both of the moment of inertia to find the value of the assumed radius.
Complete step by step solution:
We know that the moment of inertia of a hollow cylinder with inner and outer radius about its axis can be written as
$I = m ( \dfrac{R_i ^2 + R_o ^2}{2} )$
Where I is the moment of inertia of the cylinder.
The mass of the cylinder denoted by m
And the inner and outer radius are denoted by $R_i$ and $R_o$
Now we are given the inner radius of the cylinder as 10 cm
The outer radius of the cylinder as 20cm
The moment of inertia of the hollow cylinder is given as I
So we can find the moment of inertia of the cylinder about its axis as
$I = m ( \dfrac{10^2 + 20^2}{2} )$
Now we have found the moment of inertia of the hollow cylinder and let us know to find the moment of inertia of a thin cylinder with no inner and outer radius.
We know the moment of inertia of thin cylinder as : $I = mk^2$
Where k is the radius of the cylinder and I is the moment of inertia of hollow cylinder
Now we are given that the moment of inertia of the two cylinders are equal and hence we can obtain an equation as below : $I = m ( \dfrac{10^2 + 20^2}{2} ) = mk^2$
Using the above equation we can get the value of the assumed radius of the hollow cylinder as
$k = \sqrt{\dfrac{10^2 + 20^2}{2} }$
$k = 5\sqrt{10} = 16$ approx
Hence we have found the value of the radius of the cylinder as: 16 cm
Note: One of the possible mistakes that we can make in this kind of problem is that we may consider the formula for the hollow cylinder and the cylinder with thin radius as the same which is wrong. We need to consider the average of the squared inner and outer radius as the effective radius for the hollow cylinder with inner and outer radius.
Complete step by step solution:
We know that the moment of inertia of a hollow cylinder with inner and outer radius about its axis can be written as
$I = m ( \dfrac{R_i ^2 + R_o ^2}{2} )$
Where I is the moment of inertia of the cylinder.
The mass of the cylinder denoted by m
And the inner and outer radius are denoted by $R_i$ and $R_o$
Now we are given the inner radius of the cylinder as 10 cm
The outer radius of the cylinder as 20cm
The moment of inertia of the hollow cylinder is given as I
So we can find the moment of inertia of the cylinder about its axis as
$I = m ( \dfrac{10^2 + 20^2}{2} )$
Now we have found the moment of inertia of the hollow cylinder and let us know to find the moment of inertia of a thin cylinder with no inner and outer radius.
We know the moment of inertia of thin cylinder as : $I = mk^2$
Where k is the radius of the cylinder and I is the moment of inertia of hollow cylinder
Now we are given that the moment of inertia of the two cylinders are equal and hence we can obtain an equation as below : $I = m ( \dfrac{10^2 + 20^2}{2} ) = mk^2$
Using the above equation we can get the value of the assumed radius of the hollow cylinder as
$k = \sqrt{\dfrac{10^2 + 20^2}{2} }$
$k = 5\sqrt{10} = 16$ approx
Hence we have found the value of the radius of the cylinder as: 16 cm
Note: One of the possible mistakes that we can make in this kind of problem is that we may consider the formula for the hollow cylinder and the cylinder with thin radius as the same which is wrong. We need to consider the average of the squared inner and outer radius as the effective radius for the hollow cylinder with inner and outer radius.
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