# System of Particles and Rotational Motion - Class 11 - Physics

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## Extra Questions - System of Particles and Rotational Motion | NCERT | Physics | Class 11

A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system calculated about $O$ and $P$ are denoted by $\vec{L}_O$ and $\vec{L}_P$ respectively, then:

Options:

$\vec{L}{O}$ and $\vec{L}{P}$ do not vary with time

$\vec{L}{O}$ varies with time while $\vec{L}{P}$ remains constant

$\vec{L}{O}$ remains constant while $\vec{L}{P}$ varies with time

$\vec{L}{O}$ and $\vec{L}{P}$ both vary with time

To solve this problem, we start by understanding the concept of **angular momentum** in circular motion and how it changes when calculated from different points.

**Understanding Angular Momentum ($\vec{L}$):**Angular momentum $\vec{L}$ of a mass $m$ moving at a velocity $\vec{v}$ and positioned at $\vec{r}$ from a point about which the angular momentum is being calculated is given by: $$ \vec{L} = \vec{r} \times m\vec{v} $$**Angular Momentum about O ($\vec{L}_O$):**Here, $\vec{r}_O$ is the position vector from point $O$ to the mass $m$. Since the mass $m$ is undergoing circular motion about $O$ with a constant radius $R$, the radius vector $\vec{r}_O$ does not change in magnitude, only in direction.

The mass $m$ has a constant speed ($v = R\omega$) where $\omega$ is the angular velocity. As $\omega$ is constant and $O$ is the center of the circle, the velocity vector $\vec{v}$ is always tangent to the circle.

Since $\vec{v}$ is perpendicular to $\vec{r}_O$, the magnitude of angular momentum about $O$ stays constant: $$ L_O = |\vec{r}_O| \cdot m \cdot |\vec{v}| = R \cdot m \cdot R\omega = mR^2\omega $$

Since $R$ and $\omega$ are constants, $\vec{L}_O$ is constant in magnitude and because $\vec{r}_O$ and $\vec{v}$ are always perpendicular, the direction of $\vec{L}_O$ remains perpendicular to the plane of motion. Therefore, $\vec{L}_O$ remains constant over time.

**Angular Momentum about P ($\vec{L}_P$):**Here, $\vec{r}_P$ is the position vector from point $P$ to the mass $m$. This vector changes in both magnitude and direction as $m$ moves because $P$ is not at the center of the circle. It has an additional horizontal component contributing to changes in $\vec{r}_P$.

The velocity vector $\vec{v}$ remains tangent to the circle.

Since both the magnitude and direction of $\vec{r}_P$ are changing, and since $\vec{v}$ is also changing relative to $\vec{r}_P$ (not just in direction but also in angle of intersection), $\vec{L}_P = \vec{r}_P \times m\vec{v}$ varies in both magnitude and direction. This means that angular momentum about $P$ is not constant.

**Conclusion: **Based on the information and analysis above, the response to the question is: 3) $\vec{L}{O}$ remains constant while $\vec{L}{P}$ varies with time

The constant nature of $\vec{L}{O}$ arises from the constant distance of $m$ from $O$ and constant angular speed $\omega$, while the varying nature of $\vec{L}{P}$ results from the changing position vector $\vec{r}_P$ as the mass moves along its circular path.

A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x$-$y$ plane with center at $O$ and constant angular speed $\omega$. If the angular momentum of the system calculated about $O$ and $P$ are denoted by $\vec{L_O}$ and $\vec{L_P}$ respectively, then.

Options:
1): $\vec{L_O}$ and $\vec{L_P}$ do not vary with time

2): $\vec{L_O}$ varies with time while $\vec{L_P}$ remains constant

3): $\vec{L_O}$ remains constant while $\vec{L_P}$ varies with time

4): $\vec{L_O}$ and $\vec{L_P}$ both vary with time

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A thin square plate of side $3 \mathrm{~m}$ has mass $5 \mathrm{~kg}$. Find the moment of inertia about axis $AB$ as shown in the figure.

(A) $15 \mathrm{~kg}\text{-m}^{2}$

B $30 \mathrm{~kg}\text{-m}^{2}$

C $45 \mathrm{~kg}\text{-m}^{2}$ (D) $7.5 \mathrm{~kg}\text{-m}^{2}$

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Three identical stars of mass $m$ each are located at the corners of an equilateral triangle of side $L$ and are orbiting under mutual gravitational force. The orbital speed of such a 3-star system is $V_{o}=\sqrt{\frac{p G m}{L}}$. What is the value of $p$?

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A ball of mass $20 \ \mathrm{kg}$ moving at $2 \ \mathrm{m/s}$ horizontally strikes another ball of mass $12 \ \mathrm{kg}$ at rest. If after the collision, the $20 \ \mathrm{kg}$ ball is now moving with a velocity of $1 \ \mathrm{m/s}$ at an angle of $30^\circ$ with the horizontal, find the final velocity of the $12 \ \mathrm{kg}$ ball.

(A) $2.54\ \mathrm{i} + 1.62\ \mathrm{j}$

(B) $2.54\ \mathrm{i} - 1.62\ \mathrm{j}$

(C) $1.89\ \mathrm{i} + 0.83\ \mathrm{j}$

(D) $1.89\ \mathrm{i} - 0.83\ \mathrm{j}$

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Two rods of equal lengths (l) and equal mass $m$ are kept along x and y-axis, respectively, such that their centre of mass lies at the origin. The moment of inertia about a line $y=x$ is

(A) $\frac{m^{2}}{3}$

B $\frac{m l^{2}}{4}$

C $\frac{m l^{2}}{12}$

D $\frac{m l^{2}}{6}$

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Angular velocities of two bodies rotating in the same direction are given by $\omega_{1} = 30 \text{ rad/sec}$ and $\omega_{2} = 60 \text{ rad/sec}$. The relative angular velocity of the $2^{nd}$ body with respect to the $1^{st}$ body is:

A) $30 \text{ rad/sec}$ B) $40 \text{ rad/sec}$ C) $50 \text{ rad/sec}$ D) $60 \text{ rad/sec}$