1.
A thin uniform circular ring is rolling down an inclined plane of inclination $${30^ \circ }$$ without slipping. Its linear acceleration along the inclined plane will be
Acceleration of the centre of mass of the rolling body is given by $$a = \frac{{g\sin \theta }}{{1 + \left( {\frac{I}{{M{R^2}}}} \right)}}$$
Moment of inertia of the ring about an axis perpendicular to the plane of the ring and passing through its centre is given by $$I = M{R^2}$$
$$\eqalign{
& \therefore a = \frac{{g\sin \theta }}{{1 + \frac{{M{R^2}}}{{M{R^2}}}}} \cr
& = \frac{{g\sin {{30}^ \circ }}}{{1 + 1}} \cr
& = \frac{g}{4} \cr} $$
2.
Point masses $$1,2,3$$ and $$4\,kg$$ are lying at the points $$\left( {0,0,0} \right),\left( {2,0,0} \right),\left( {0,3,0} \right)$$ and $$\left( { - 2, - 2,0} \right)$$ respectively. The moment of inertia of this system about $$X$$-axis will be
3.
A circular disc $$X$$ of radius $$R$$ is made froth an iron plate of thickness $$t ,$$ and another disc $$Y$$ of radius $$4R$$ is made from an iron plate of thickness $$\frac{t}{4}.$$ Then the relation between the moment of inertia $${I_X}$$ and $${I_Y}$$ is-
We know that density $$\left( d \right) = \frac{{{\text{mass}}\left( M \right)}}{{{\text{volume}}\left( V \right)}}$$
$$\therefore M = d \times V = d \times \left( {\pi {R^2} \times t} \right)$$
The moment of inertia of a disc is given by $$I = \frac{1}{2}M{R^2}$$
4.
A cubical block of side a is moving with velocity $$V$$ on a horizontal smooth plane as shown in Figure. It hits a ridge at point $$O.$$ The angular speed of the block after it hits $$O$$ is
A.
$$\frac{{3V}}{{\left( {4a} \right)}}$$
B.
$$\frac{{3V}}{{\left( {2a} \right)}}$$
C.
$$\frac{{\sqrt {3V} }}{{\left( {\sqrt {2a} } \right)}}$$
$$r = \sqrt 2 \frac{a}{2}\,\,\,\,\,\,or,\,{r^2} = \frac{{{a^2}}}{2}$$
Net torque about $$O$$ is zero.
Therefore, angular momentum $$\left( L \right)$$ about $$O$$ will be conserved, or $${L_i} = {L_f}$$
$$\eqalign{
& MV\left( {\frac{a}{2}} \right) = {I_0}\,\omega = \left( {{I_{cm}} + M{r^2}} \right)\omega \cr
& \omega = \left\{ {\frac{{M{a^2}}}{6} + M\left( {\frac{{{a^2}}}{2}} \right)} \right\}\omega = \frac{2}{3}M{a^2}\omega = \frac{{3v}}{{4a}} \cr} $$
5.
Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is $$m.$$ The mass of the ink used to draw the outer circle is $$6 \,m.$$
The coordinates of the centres of the different parts are: outer circle $$\left( {0,0} \right),$$ left inner circle $$\left( { - a,a} \right),$$ right inner circle $$\left( {a,a} \right),$$ vertical line $$\left( {0,0} \right)$$ and horizontal line $$\left( {0, - a} \right).$$ The $$y$$-coordinate of the centre of mass of the ink in this drawing is
The system is made up of five bodies (three circles and two straight lines) of uniform mass distribution. Therefore we assume the system to be made up of five point masses where the mass of each body is considered at its geometrical centre.
The y-coordinate of the centre of mass is
$$\eqalign{
& {y_{cm}} = \frac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3} + {m_4}{y_4} + {m_5}{y_5}}}{{{m_1} + {m_2} + {m_3} + {m_4} + {m_5}}} \cr
& \therefore {y_{cm}} = \frac{{6m \times 0 + m \times 0 + m \times a + m \times a + m\left( { - a} \right)}}{{6m + m + m + m + m}} \cr
& = \frac{{ma}}{{10m}} \cr
& = \frac{a}{{10}} \cr} $$
6.
Angular momentum of the particle rotating with a central force is constant due to-
We know that $$\overrightarrow {{\tau _c}} = \frac{{d\overrightarrow {{L_c}} }}{{dt}}$$
Where $$\overrightarrow {{\tau _c}} = $$ torque about the center of mass of the body and $$\overrightarrow {{L_c}} = $$ Angular momentum about the center of mass of the body. Central forces act along the center of mass.
Therefore torque about center of mass is zero.
When $$\overrightarrow {{\tau _c}} = 0$$ then $$\overrightarrow {{L_c}} = {\text{constant}}$$
7.
A thin uniform rod, pivoted at $$O,$$ is rotating in the horizontal plane with constant angular speed $$\omega ,$$ as shown in the figure. At time $$t=0,$$ a small insect starts from $$O$$ and moves with constant speed $$v,$$ with respect to the rod towards the other end. It reaches the end of the rod at $$t=T$$ and stops. The angular speed of the system remains $$\omega $$ throughout. The magnitude of the torque $$\left( {\left| {\vec \tau } \right|} \right)$$ about $$O,$$ as a function of time is best represented by which plot?
We know that $$\left| {\vec \tau } \right| = \left| {\frac{{d\vec L}}{{dt}}} \right|$$ where $$L = I\omega $$
$$\therefore \,\,\tau = \frac{d}{{dt}}\left( {I\omega } \right) = \omega \frac{{dI}}{{dt}}\,.....(i)$$
From the situation it is clear that the moment of inertia for (rod + insect) system is increasing.
Let at any instant of time $$'t ’,$$ the insect is at a distance $$x$$ from $$O.$$ At this instant, the moment of inertia of the system is
$$\eqalign{
& I = \frac{1}{3}M{L^2} + m{x^2}\,\,.....(ii) \cr
& {\text{From }}\,(i)\,\,\& \,\,(ii) \cr
& \tau = \omega \frac{d}{{dt}}\left[ {\frac{1}{3}M{L^2} + m{x^2}\,} \right] = \omega m\frac{d}{{dt}}\left( {{x^2}} \right) \cr
& = 2\omega mx\frac{{dx}}{{dt}} \cr
& = 2\omega mxv\,\,\,\,\,\left[ {\because x = vt} \right] \cr
& \therefore \tau \propto t\,\,\,\,\,\left( {{\text{till }}t = T} \right) \cr} $$
When the insect stops moving, $${\vec L}$$ does not change and therefore $$\tau $$ becomes constant.
8.
An automobile moves on a road with a speed of $$54\,km{h^{ - 1}}.$$ The radius of its wheels is $$0.45\,m$$ and the moment of inertia of the wheel about its axis of rotation is $$3\,kg\,{m^2}.$$ If the vehicle is brought to rest in $$15\,s,$$ the magnitude of average torque transmitted by its brakes to the wheel is
Vertical line from hinge $$A$$ must pass through $$C.M.$$ of rod system.
$$\eqalign{
& \tan \theta = \frac{{OP}}{{AP}} = \frac{{\frac{\ell }{2}}}{{\frac{{2\ell }}{3}}} \cr
& \tan \theta = \frac{3}{4} \Rightarrow \theta = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right) \cr} $$
10.
A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?
The net force acting on a particle undergoing uniform circular motion is centripetal force which always passes through the centre of the circle. The torque due to this
force about the centre is zero, therefore, $${\vec L}$$ is conserved about $$O.$$