Question 101:
easyA 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 $\alpha = \frac{F}{MR} = \frac{30}{3 \times 0.4} = 25 \text{ rad/s}^2$.
Question 102:
easyTwo 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 $L_B > L_A$.
Question 103:
easyAn 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 $\tau = I\alpha = 3 \times (\frac{20}{9}) = 6.66 \text{ kg m}^2/\text{s}^2$.
Question 104:
easyA 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)
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 $\alpha = -1$.
Question 105:
moderateA solid cylinder of mass $50\text{ kg}$ and radius $0.5\text{ m}$ is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of $2\text{ rev/s}^{2}$ is:
(2014)
Moment of inertia $I = \frac{1}{2}MR^{2} = 6.25\text{ kg m}^{2}$. Angular acceleration $\alpha = 2\text{ rev/s}^{2} = 4\pi\text{ rad/s}^{2}$. Torque $\tau = I\alpha = 25\pi = 78.5\text{ N m}$. Tension $T = \frac{\tau}{R} = \frac{78.5}{0.5} = 157\text{ N}$.
Question 106:
easyA circular platform is mounted on a frictionless vertical axle. Its radius $R = 2\text{ m}$ and its moment of inertia about the axle is $200\text{ kg m}^{2}$. It is initially at rest. A $50\text{ kg}$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1\text{ ms}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is:
(2012 Mains)
Angular velocity of man $\omega_{m} = \frac{v}{R} = 0.5\text{ rad/s}$. By conservation of angular momentum, $I_{p}\omega_{p} + I_{m}\omega_{m} = 0$, giving $\omega_{p} = -0.5\text{ rad/s}$. Relative angular velocity $\omega_{rel} = \omega_{m} - \omega_{p} = 1\text{ rad/s}$. Time $T = \frac{2\pi}{\omega_{rel}} = 2\pi\text{ s}$.
Question 107:
easyWhen 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 $\vec{p}$.
Question 108:
moderateThe moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its midpoint and perpendicular to its length is $I_{0}$. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is
(2011 Mains)
Using the parallel axis theorem, $I = I_{cm} + Md^{2}$. Here, the center of mass moment of inertia is $I_{cm} = I_{0}$ and the distance to the parallel axis is $d = \frac{L}{2}$. Thus, $I = I_{0} + M(\frac{L}{2})^{2} = I_{0} + \frac{ML^{2}}{4}$.
Question 109:
easyThe instantaneous angular position of a point on a rotating wheel is given by the equation $$\theta (t) = 2t^{3} – 6t^{2}$$. The torque on the wheel becomes zero at:
(2011 Pre)
Angular velocity $\omega = \frac{d\theta}{dt} = 6t^{2} - 12t$. Angular acceleration $\alpha = \frac{d\omega}{dt} = 12t - 12$. Torque is zero when $\alpha = 0$, which implies $12t - 12 = 0$, or $t = 1\text{ s}$.
Question 110:
moderateFrom a circular disc of radius $R$ and mass $9M$, a small disc of mass $M$ and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its center is:
(2010 Mains)
Initial moment of inertia $I_{original} = \frac{1}{2}(9M)R^{2} = \frac{9}{2}MR^{2}$. Moment of inertia of the removed part is $I_{removed} = \frac{1}{2}M(\frac{R}{3})^{2} = \frac{1}{18}MR^{2}$. The remaining moment of inertia is $I = I_{original} - I_{removed} = \frac{9}{2}MR^{2} - \frac{1}{18}MR^{2} = \frac{40}{9}MR^{2}$.