Question 705022

SCQHARD

A uniform plane square sheet of mass $m$ is centered at the origin of an inertial frame. The sheet is rotating about an axis passing through the origin. At an instant when all its vertices lie on $x$ and $y$ axes, the angular momentum is $\vec{L} = I_0\omega_0(2\hat{i} + \hat{j} + 2\hat{k})$ , where $I_0$ is the moment of inertia about the $x$ axis. At this instant, the angular velocity of the sheet is

(A)

$(2\hat{i} + \hat{j} + 2\hat{k})\omega_0$

(B)

$(2\hat{i} + \hat{j} + \hat{k})\omega_0$

(C)

$(2\hat{i} + \hat{j})\omega_0$

(D)

$(\hat{i} + \hat{j})\omega_0$

Detailed Solution

For a square sheet of mass $m$ , the principal moments of inertia are $I_{xx} = I_{yy} = \frac{ma^2}{12}$ and $I_{zz} = \frac{ma^2}{6}$ .

Setting $I_0 = \frac{ma^2}{12}$ , the inertia tensor in the given frame is diagonal: $I = diag(I_0, I_0, 2I_0)$ .

Using $\vec{L} = I\vec{\omega}$ , where $\vec{L} = I_0\omega_0(2, 1, 2)$ , we have $I_0\omega_0(2, 1, 2) = I_0 \omega_x \hat{i} + I_0 \omega_y \hat{j} + 2I_0 \omega_z \hat{k}$ . Comparing components: $\omega_x = 2\omega_0$ , $\omega_y = \omega_0$ , and $2\omega_z = 2\omega_0 \implies \omega_z = \omega_0$ . Thus, $\vec{\omega} = (2\hat{i} + \hat{j} + \hat{k})\omega_0$ .

Question 705022

Detailed Solution

Boost Your Exam Preparation!