Question

Comprehension Type Questions Paragraph for Q. 33 to Q. 34 In the treatment of moments of inertia, introductory textbooks often present two theorems, generally called the parallel axis theorem and the perpendicular axis theorem. There is another theorem of this same genre, which is not usually included, but which is interesting and useful. It is $I_x + I_y + I_z = 2 \sum m_i r_i^2$ Here, $I_x$, $I_y$ and $I_z$ are the moments of inertia about three mutually perpendicular intersecting axes, $m_i$ is the mass of the $i^{th}$ particle and $r_i$ is the distance from the intersection. The proof is simple: Taking the three axes as coordinate axes, we have: $I_x + I_y + I_z$ $= \sum m_i (y_i^2 + z_i^2) + \sum m_i (z_i^2 + x_i^2) + \sum m_i (x_i^2 + y_i^2)$ $= 2 \sum m_i (x_i^2 + y_i^2 + z_i^2) = 2 \sum m_i r_i^2$ One important application is the calculation of the moment of inertia $I_d$ of a uniform thin-walled spherical shell, of mass M and radius R, about a diameter. Taking the centre as the origin of coordinates, we have $I_x = I_y = I_z = I_d$, and $r_i = R$. The theorem gives $3I_d = 2 \sum m_i R^2 = 2( \sum m_i) R^2 = 2MR^2$, where $I_d = \frac{2MR^2}{3}$. 33. Consider a solid cube of mass m and side L. What will be the value of $\sum m_i r_i^2$ for this body when the point of intersection of axes is the centre of the cube: (A) $\frac{ML^2}{2}$ (B) $\frac{ML^2}{4}$ (C) $\frac{ML^2}{3}$ (D) $\frac{ML^2}{6}$ 34. Find the moment of inertia of ring of mass m and radius R about an axis passing through its centre and making an angle 45° with its plane: (A) $\frac{MR^2}{4}$ (B) $\frac{MR^2}{2}$ (C) $\frac{3}{4} MR^2$ (D) $MR^2$

          Comprehension Type Questions
Paragraph for Q. 33 to Q. 34
In the treatment of moments of inertia, introductory textbooks often present two theorems, generally called the parallel axis theorem and the perpendicular axis theorem. There is another theorem of this same genre, which is not usually included, but which is interesting and useful.
It is
$I_x + I_y + I_z = 2 \sum m_i r_i^2$
Here, $I_x$, $I_y$ and $I_z$ are the moments of inertia about three mutually perpendicular intersecting axes, $m_i$ is the mass of the $i^{th}$ particle and $r_i$ is the distance from the intersection. The proof is simple: Taking the three axes as coordinate axes, we have:
$I_x + I_y + I_z$
$= \sum m_i (y_i^2 + z_i^2) + \sum m_i (z_i^2 + x_i^2) + \sum m_i (x_i^2 + y_i^2)$
$= 2 \sum m_i (x_i^2 + y_i^2 + z_i^2) = 2 \sum m_i r_i^2$
One important application is the calculation of the moment of inertia $I_d$ of a uniform thin-walled spherical shell, of mass M and radius R, about a diameter. Taking the centre as the origin of coordinates, we have $I_x = I_y = I_z = I_d$, and $r_i = R$. The theorem gives $3I_d = 2 \sum m_i R^2 = 2( \sum m_i) R^2 = 2MR^2$, where $I_d = \frac{2MR^2}{3}$.
33. Consider a solid cube of mass m and side L. What will be the value of $\sum m_i r_i^2$ for this body when the point of intersection of axes is the centre of the cube:
(A) $\frac{ML^2}{2}$ (B) $\frac{ML^2}{4}$ (C) $\frac{ML^2}{3}$ (D) $\frac{ML^2}{6}$
34. Find the moment of inertia of ring of mass m and radius R about an axis passing through its centre and making an angle 45° with its plane:
(A) $\frac{MR^2}{4}$ (B) $\frac{MR^2}{2}$ (C) $\frac{3}{4} MR^2$ (D) $MR^2$

comprehension type questions paragraph for q 33 to q 34 in the treatment of moments of inertia introductory textbooks often present two theorems generally called the parallel axis theorem an 43786

Added by Matthew N.

Question

Please give Ace some feedback