A light rod of length $\ell$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:
(2016 – II)
The moment of inertia about the center of mass uses reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$, resulting in $I = \mu \ell^2 = \frac{m_1 m_2}{m_1 + m_2}\ell^2$.
A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^\circ$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is: (2008)
Treat the bent rod as two rods of length $L/2$ and mass $M/2$ connected at the origin. Using the moment of inertia formula for a rod about its end $I = \frac{m l^2}{3}$, total $I = 2 \times \frac{(M/2)(L/2)^2}{3} = \frac{ML^2}{12}$.
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is:
(2004)
For disc about in-plane tangent, $I_d = \frac{5}{4}MR^2 \implies k_d = \frac{\sqrt{5}}{2}R$. For ring about in-plane tangent, $I_r = \frac{3}{2}MR^2 \implies k_r = \sqrt{\frac{3}{2}}R$. Ratio is $\sqrt{5}:\sqrt{6}$.
A circular disc is to be made by using iron and aluminium so that it acquired maximum moment of inertia about geometrical axis. It is possible with:
(2002)
Moment of inertia is $I = \int r^2 dm$. To maximize $I$, denser material (iron) should be placed at the outer periphery and lighter material (aluminium) at the interior.
A solid spherical ball rolls on a table. Ratio of its rotational kinetic energy to total kinetic energy is:
(1994)
For a solid sphere, $I = \frac{2}{5}MR^2$. Rotational kinetic energy is $\frac{1}{2}Iomega^2 = \frac{1}{5}MR^2\omega^2$ and total kinetic energy is $\frac{7}{10}MR^2\omega^2$. Ratio is $\frac{2}{7}$.
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)
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.
The moment of the force, $\vec{F} = 4\hat{i} + 5\hat{j} – 6\hat{k}$ at $(2, 0, -3)$, about the point $(2, -2, -2)$ is given by
(2018)
The relative position vector is $\vec{r} = (2-2)\hat{i} + (0 - (-2))\hat{j} + (-3 - (-2))\hat{k} = 2\hat{j} - \hat{k}$. Torque is $\vec{\tau} = \vec{r} \times \vec{F} = (2\hat{j} - \hat{k}) \times (4\hat{i} + 5\hat{j} - 6\hat{k}) = -7\hat{i} - 4\hat{j} - 8\hat{k}$.
A rope is wound around a hollow cylinder of mass $3 \text{ kg}$ and radius $40 \text{ cm}$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 \text{ N}$?
(2017-Delhi)
For a hollow cylinder, the moment of inertia is $I = MR^2$. The torque provided by the rope is $\tau = F \times R = I\alpha$. Substituting $I$, we get $F \times R = MR^2 \alpha$, which gives $\alpha = \frac{F}{MR} = \frac{30}{3 \times 0.4} = 25 \text{ rad/s}^2$.
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)
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 $L_B > L_A$.
An automobile moves on a road with a speed of $54 \text{ km/h}$. The radius of its wheels is $0.45 \text{ m}$ and the moment of inertia of the wheel about its axis of rotation is $3 \text{ kgm}^2$. If the vehicle is brought to rest in $15 \text{ s}$, the magnitude of average torque transmitted by its brakes to wheel is:
(2015 Re)
Initial angular velocity $\omega_0 = \frac{v}{r} = \frac{15}{0.45} = \frac{100}{3} \text{ rad/s}$. The angular acceleration is $\alpha = \frac{\omega_0}{t} = \frac{100/3}{15} = \frac{20}{9} \text{ rad/s}^2$. Torque is $\tau = I\alpha = 3 \times (\frac{20}{9}) = 6.66 \text{ kg m}^2/\text{s}^2$.