Question

please solve this physics question

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,

The following question reflects a bit of parallel axis theorem. Parallel axis theorem is stating the particular rigid body containing a moment of inertia for any sides of axis will be equivalent to moment of inertia in a "parallel axis" which is definitely crossing over the centre of mass including its product of mass of it's body and a square of that perpendicular distance about those two axes.

We can conclude that by this formulae,,,

$\boxed{\bf{I = I_{cm} + Mh^2}}$

OR

$\boxed{\bf{I = I_{cm} + MD^2}}$

Here, the letters can be denoted as, $\textbf{"I"}$ is denoting the moment of inertia about a certain axis in the given cases, $\bf{"I_{cm}"}$ denotes the axial passages through a given centre of mass, $\textbf{"M"}$ is a total attained and present mass of the whole body and $\textbf{"h" OR "H" OR "d" OR "D"}$ is representing a particular distance maintained between those two axes.

Now, we are given a figure with an axis of $\textbf{"O"}$, therefore the equation of parallel theorem becomes such like this ..

$\boxed{\bf{I_O = I_{cm} + Mh^2}}$

Now, here, $\bf{"I"}$ is denoting the moment of inertia about an $\textbf{Rotational Axis "O"}$ with a certain distance as $\bf{"H"}$ from it's object's centre of mass in it, $\bf{"I_{cm}"}$ denotes the axial passage about an axis of "O" through a given centre of mass and $\bf{"M"}$ is a total attained and present mass of the whole body or is the rigid object's total mass.

By applying that theorem of parallel axis and imply the given data to us by the figure to find the moment for inertia,, here,, we get ...

$\boxed{\bf{I_O = I_{cm} + Mh^2}}$

Note that; the circles are "7" in number and, axis passing through the circle gives a moment of inertia half of that in respect to mass and radius since, it'll get divided into two.. and one circular point divides those "6" circles by a single point of "O" giving us the radius as "2R" ... therefore,,,,

$\bf{I_O = \frac{7MR^2}{2} + 6 \times (M \times (2R)^2)}$

$\bf{I_O = \frac{7MR^2}{2} + 2^2 \times 6MR^2}$

$\bf{I_O = \frac{7MR^2}{2} + \frac{6M \times 2^2 R^2 \times 2}{2}}$

$\bf{I_O = \frac{7MR^2}{2} + \frac{2^2 \times 12MR^2}{2}}$

$\bf{I_O = \frac{7MR^2}{2} + \frac{48MR^2}{2}}$

$\bf{I_O = \frac{55MR^2}{2}}$

Another system of it'll be representing here, $\bf{"I"}$ is denoting the moment of inertia about an $\textbf{Rotational Axis "P"}$ with a certain distance as $\bf{"H"}$ from it's object's centre of mass in it, $\bf{"I_{cm}"}$ denotes the axial passage about an axis of "P" through a given centre of mass and $\bf{"M"}$ is a total attained and present mass of the whole body or is the rigid object's total mass.

By applying that theorem of parallel axis and imply the given data to us by the figure to find the moment for inertia,, here,, we get ...
Moment of inertia's second system with that axis of Point "P"...

$\bf{I_P = I_{cm} + Mh^2}$

$\bf{I_P = \frac{55MR^2}{2} + 7M \: (3R)^2}$

$\bf{I_P = \frac{55MR^2}{2} + \frac{7M \times 3^2 R^2 \times 2}{2}}$

$\bf{I_P = \frac{55MR^2 + (3^2 \times 14MR^2)}{2}}$

$\bf{I_P = \frac{55MR^2 + 126MR^2}{2}}$

$\bf{I_P = \frac{181MR^2}{2}}$

From the given option in the query,,, answer is,,,

$\boxed{\huge{\bf{Option \: c) \: \frac{181MR^2}{2}}}}$

HOPE THIS DETAILED ANSWER HELPS YOU AND EASILY SOLVES YOUR DOUBTS FOR CALCULATING MOMENT OF INERTIA!!!!