When a mass is rotating in a plane about a fixed point, its angular momentum is directed along
(2012 Pre)
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)
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)
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)
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)
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)
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)
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 $\frac{1}{8}I(\omega_1-\omega_2)^2$.
A planet is moving in an elliptical orbit around the sun. If $T$, $V$, $E$ and $L$ stand respectively for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?
(1990)
For a bound elliptical orbit, total energy $E$ is always negative. $T$ and $V$ vary with distance, and $L$ is conserved in both magnitude and direction as Torque is Zero. Gravitational force is passing through Center of Rotation so Toque is zero.