A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
(2018)
1. Rotational kinetic energy
2. Moment of inertia
3. Angular velocity
4. Angular momentum
View Answer
In the absence of any external torque acting on the sphere in free space, the angular momentum of the system remains conserved. Therefore, angular momentum remains constant while moment of inertia increases and angular velocity decreases.
Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B$ ($I_B > I_A$) have equal kinetic energy of rotation. If $L_A$ and $L_B$ be their angular momenta respectively, then:
(2016-II)
1. $L_B > L_A$
2. $L_A > L_B$
3. $L_A = \frac{L_B}{2}$
4. $L_A = 2L_B$
View Answer
Rotational kinetic energy is related to angular momentum by the formula $K = \frac{L^2}{2I}$, which yields $L = \sqrt{2KI}$. Since $K$ is the same for both bodies and it is given that $I_B > I_A$, it directly follows that $L_B > L_A$.
A force $\vec{F} = \alpha\hat{i} + 3\hat{j} + 9\hat{k}$ is acting at a point $\vec{r} = 2\hat{i} – 6\hat{j} – 12\hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is: (2015 Re)
1. $1$
2. $-1$
3. $2$
4. Zero
View Answer
For angular momentum to be conserved, torque $\vec{\tau} = \vec{r} \times \vec{F}$ must be zero, meaning $\vec{r}$ and $\vec{F}$ are collinear. Taking the ratio of their components: $\frac{2}{\alpha} = \frac{-6}{3} = \frac{-12}{9}$, which simplifies to $\frac{2}{\alpha} = -2$, giving $\alpha = -1$.
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along
(2012 Pre)
1. A line perpendicular to the plane of rotation
2. The line making an angle of $45^{\circ}$ to the plane of rotation
3. The radius
4. The tangent to the orbit
View Answer
Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$. According to the properties of the cross product, the vector $\vec{L}$ is directed perpendicular to the plane containing the position vector $\vec{r}$ and momentum vector $\vec{p}$.
A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with constant angular velocity $\omega$. The objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by:
(2010 Mains)
1. $\frac{(M+2m)\omega}{2m}$
2. $\frac{2M\omega}{M+2m}$
3. $\frac{(M+2m)\omega}{M}$
4. $\frac{M\omega}{M+2m}$
View Answer
By conservation of angular momentum, $I_{1}\omega_{1} = I_{2}\omega_{2}$. Initially, $I_{1} = Mr^{2}$. Finally, the moment of inertia is $I_{2} = Mr^{2} + 2mr^{2} = (M+2m)r^{2}$. Equating the two yields $Mr^{2}\omega = (M+2m)r^{2}\omega_{2}$, so $\omega_{2} = \frac{M\omega}{M+2m}$.
A thin circular ring of mass $M$ and radius ‘$r$’ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:
(2003)
1. $\frac{M\omega}{4m}$
2. $\frac{M\omega}{M + 4m}$
3. $\frac{(M + 4m)\omega}{M}$
4. $\frac{(M + 4m)\omega}{M + 4m}$
View Answer
By conservation of angular momentum, $I_{initial}\omega_{initial} = I_{final}\omega_{final}$.
Initial moment of inertia $I_i = Mr^2$. Final moment of inertia $I_f = Mr^2 + 4mr^2$.
Thus, $$Mr^2 \omega = (M + 4m)r^2 \omega' \implies \omega' = \frac{M\omega}{M + 4m}$$.
A disc is rotating with angular speed $\omega$. If a child sits on it, what is conserved:
(2002)
1. Linear momentum
2. Angular momentum
3. Kinetic energy
4. Potential energy
View Answer
When the child sits on the rotating disc gently, no external torque acts on the system.
According to Newton's second law for rotation, if net external torque is zero, the total angular momentum of the system remains conserved.
A circular ring of mass $M$ and radius $R$ is rotating about its axis with constant angular velocity $\omega$. Two particles each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The angular velocity of the ring will now become:
(1998)
1. $\frac{M - 2m}{m\omega}$
2. $\frac{m\omega}{M - 2m}$
3. $\frac{M\omega}{M + 2m}$
4. $\frac{M - 2m}{m}$
View Answer
Since the particles are attached gently, external torque is zero, meaning angular momentum is conserved.
$I_1\omega_1 = I_2\omega_2 \implies (MR^2)\omega = (MR^2 + 2mR^2)\omega'$.
Solving for the new angular velocity yields $\omega' = \frac{M\omega}{M + 2m}$.
A particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt{2}$ in the XOY plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is:
(1991)
1. $60 \text{ units}$
2. $40\sqrt{2} \text{ units}$
3. Zero
4. $7.5 \text{ units}$
View Answer
Angular momentum $L = mvr_{\perp}$. The line equation is $X - Y + 4 = 0$.
The perpendicular distance $r_{\perp}$ from the origin $(0,0)$ to the line is $\frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Thus, $L = 5 \times (3\sqrt{2}) \times (2\sqrt{2}) = 60 \text{ units}$.
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and $\omega_2$. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:
(2017-Delhi)
1. $\frac{1}{4}I(\omega_1-\omega_2)^2$
2. $I(\omega_1-\omega_2)^2$
3. $\frac{1}{8}I(\omega_1-\omega_2)^2$
4. $\frac{1}{2}I(\omega_1-\omega_2)^2$
View Answer
By conservation of angular momentum, the final common angular velocity is $\omega = \frac{\omega_1 + \omega_2}{2}$. The loss in rotational kinetic energy is $Delta E = E_i - E_f = \frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2 - 2 \cdot \left(\frac{1}{2}I\omega^2\right)$, which simplifies to $\frac{1}{8}I(\omega_1-\omega_2)^2$.