Four point masses, each of mass m, are fixed at the corners of a square of side l . The square is rotating with angular frequency \omega , about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is :

Rotational Motion PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q55 with Solution

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Correct answer: A

Interpret the axis

The axis passes through one corner of the square and is parallel to a diagonal of the square.

Let the square lie in the $xy$ -plane with corners at

A(0,0),\quad B(l,0),\quad C(l,l),\quad D(0,l)

Each corner has mass $m$ .

Take the rotation axis through $A(0,0)$ and parallel to diagonal $BD$ (or equivalently $AC$ as per the figure’s intended geometry). The important point is to compute the perpendicular distance of each mass from this axis.

Moment of inertia about the given axis

If the axis is through $A$ and parallel to diagonal $BD$ , then its equation in the plane is

y=-x

(slope $-1$ , passing through origin).

The perpendicular distance of a point $(x,y)$ from line $x+y=0$ is

r_\perp = \frac{|x+y|}{\sqrt{2}}

Now compute for each corner:

For $A(0,0)$ :

r_A=0

For $B(l,0)$ :

r_B=\frac{|l+0|}{\sqrt{2}}=\frac{l}{\sqrt{2}}

For $D(0,l)$ :

r_D=\frac{|0+l|}{\sqrt{2}}=\frac{l}{\sqrt{2}}

For $C(l,l)$ :

r_C=\frac{|l+l|}{\sqrt{2}}=\sqrt{2}\,l

Hence the moment of inertia is

I = m r_A^2 + m r_B^2 + m r_C^2 + m r_D^2

I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m(\sqrt{2}l)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2

I = m\left(\frac{l^2}{2} + 2l^2 + \frac{l^2}{2}\right)=3ml^2

Angular momentum

Since the body rotates with angular frequency $\omega$ about this fixed axis,

L = I\omega = 3ml^2\omega

Match with options

So the angular momentum is

\boxed{3ml^2\omega}

which corresponds to Option A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

More from Rotational Motion

View written solutionFree

Correct answer: A

Interpret the axis

The axis passes through one corner of the square and is parallel to a diagonal of the square.

Let the square lie in the $xy$ -plane with corners at

A(0,0),\quad B(l,0),\quad C(l,l),\quad D(0,l)

Each corner has mass $m$ .

Moment of inertia about the given axis

If the axis is through $A$ and parallel to diagonal $BD$ , then its equation in the plane is

y=-x

(slope $-1$ , passing through origin).

The perpendicular distance of a point $(x,y)$ from line $x+y=0$ is

r_\perp = \frac{|x+y|}{\sqrt{2}}

Now compute for each corner:

For $A(0,0)$ :

r_A=0

For $B(l,0)$ :

r_B=\frac{|l+0|}{\sqrt{2}}=\frac{l}{\sqrt{2}}

For $D(0,l)$ :

r_D=\frac{|0+l|}{\sqrt{2}}=\frac{l}{\sqrt{2}}

For $C(l,l)$ :

r_C=\frac{|l+l|}{\sqrt{2}}=\sqrt{2}\,l

Hence the moment of inertia is

I = m r_A^2 + m r_B^2 + m r_C^2 + m r_D^2

I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m(\sqrt{2}l)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2

I = m\left(\frac{l^2}{2} + 2l^2 + \frac{l^2}{2}\right)=3ml^2

Angular momentum

Since the body rotates with angular frequency $\omega$ about this fixed axis,

L = I\omega = 3ml^2\omega

Match with options

So the angular momentum is

\boxed{3ml^2\omega}

which corresponds to Option A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

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