Rotational Motion

According to conservation of angular momentum about suspension point,

$mvx = \left( {{{m{L^2}} \over 3} + m{x^2}} \right)\omega $

$vx = \left( {{{{L^2}} \over 3} + {x^2}} \right)\omega \Rightarrow \omega = {{3vx} \over {{L^2} + 3{x^2}}}$

For maximum,

${{d\omega } \over {dx}} = 0 \Rightarrow x = {L \over {\sqrt 3 }} \Rightarrow \omega = {{\sqrt 3 v} \over {2L}}$

$\Delta $K + $\Delta $U = 0

${1 \over 2}{I_0}{\omega ^2} = - \Delta U$

${1 \over 2}{{m{l^2}} \over 3}{\omega ^2} = - \left( { - mg{l \over 4}} \right)$

$\omega = \sqrt {{{3g} \over {2l}}} $

$ \Rightarrow {a_{radial}} = {\omega ^2}{l \over 2} = {{3g} \over {2l}}{l \over 2} = {{3g} \over 4} \Rightarrow \tau = {I_0}\alpha $

$\alpha = {{mg{l \over 2}\sin 60^\circ } \over {mg{{{l^2}} \over 3}}} = {{3\sqrt 3 g} \over {4l}}$

$ \Rightarrow {a_v} = \left( {\alpha {l \over 2}} \right)\sin 60^\circ + {\omega ^2}{l \over 2}\cos 60^\circ $

${a_v} = {{3\sqrt 3 g} \over 8}{{\sqrt 3 } \over 2} + {{3g} \over 8} \Rightarrow {a_v} = {{9g} \over {16}} + {{6g} \over {16}}$

$mg - N = m{a_v} \Rightarrow N = {{mg} \over {16}}$

Given, force applied on body

$\overrightarrow F = (\alpha t\widehat i + \beta \widehat j)$

Here, $\alpha$ = 1.0 N s^$-$1 and $\beta$ = 1.0 N. Therefore,

$\overrightarrow F = t\widehat i + \widehat j$

By Newton's second law of motion, F = ma. Given m = 1.0 kg. Therefore,

$\overrightarrow F = \overrightarrow a \Rightarrow \overrightarrow a = (t\widehat i + \widehat j)$

We know $\overrightarrow a = {{d\overrightarrow v } \over {dt}} \Rightarrow \overrightarrow v = \int {\overrightarrow a dt} $

$ \Rightarrow \overrightarrow v = \int {(t\widehat i + \widehat j)dt = {{{t^2}} \over 2}\widehat i + t\widehat j} $ ..... (1)

Also, $\overrightarrow v = {{\overrightarrow {dr} } \over {dt}} \Rightarrow \overrightarrow r = \int {\overrightarrow v dt} $

$ \Rightarrow \overrightarrow r = \int {\left( {{{{t^2}} \over 2}\widehat i + t\widehat j} \right)dt = {{{t^3}} \over 6}\widehat i + {{{t^2}} \over 2}\widehat j} $ ..... (2)

Now, $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{t^3}/6} & {{t^2}/2} & 0 \cr t & 1 & 0 \cr } } \right|$

$\overrightarrow \tau = \widehat i(0) - \widehat j(0) + \widehat k\left( {{{{t^3}} \over 6} - {{{t^3}} \over 2}} \right) = {{ - 1} \over 3}{t^3}\widehat k$ N m

At t = 1.0 s, $\overrightarrow \tau = {{ - 1} \over 3}\widehat k$ N m

Hence, option (A) is correct.

Now, from Eq. (1), $\overrightarrow v = {{{t^2}} \over 2}\widehat i + t\widehat j$

At t = 1.0 s, $\overrightarrow v = \left( {{1 \over 2}\widehat i + \widehat j} \right)$ m s^$-$1 = $ = {1 \over 2}(\widehat i + 2\widehat j)$ m s^$-$1

Hence, option (C) is correct.

$V = {{k{r^2}} \over 2}$ ; where k is positive constant, r is radius of circular orbit, V(r) is potential, v is speed of particle, L is angular momentum.

$F = - {{dV} \over {dr}} = - kr$ (towards centre)



$kr = {{m{v^2}} \over R}$ (Centripetal force)

At r = R, $kR = {{m{v^2}} \over R}$

$v = \sqrt {{{k{R^2}} \over m}} = \sqrt {{k \over m}} R$

Hence, option (B) is correct.

Also, we know L = mvR

Substitute the value of v, we get

$ L = mvR = \sqrt {mk} {R^2}$

Hence, option (C) is correct

(a) If force is applied normal to surface at P, then line of action of force will pass from Q and thus $\tau$ = 0.

(b) Wheel can climb.

(c) $\tau$ = F(2Rcos$\theta$) $-$ mgRcos$\theta$

$ \Rightarrow $ $\tau$ $\propto$ cos$\theta$


Hence, as $\theta$ increases, $\tau$ decreases. So its correct.

(d)
$\tau$ = Fr_{$ \bot $} $-$ mgcos$\theta$ ; $\tau$ increases with $\theta$.

We discuss the options as follows:


$\bullet$ For option (A) : When the bar makes an angle $\theta$ with the floor, the height of the midpoint is ${L \over 2}\cos \theta $. this is height of centre of mass.

A force mg acts vertically downward. Thus, the midpoint of bar falls vertically downward. Hence, option (A) is correct.

$\bullet$ For option (B) : Here, we have $x = {L \over 2}\sin \theta $ and $y = L\cos \theta $; therefore,

$\sin \theta = {x \over {L/2}}$ and $\cos \theta = {y \over L}$

Using the condition ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we get

${{{x^2}} \over {{{(L/2)}^2}}} + {{{y^2}} \over {{L^2}}} = 1$ (equation of ellipse)

Thus, trajectory of point A is not parabola. Hence, option (B) is incorrect.

$\bullet$ For option (C) : The torque acting about the point of contact with floor, that is, at point B is

$\tau = \overrightarrow F \times \overrightarrow r = mg{L \over 2}\sin \theta $

Thus, the torque is proportional to sin$\theta$. Hence, option (C) is correct.

$\bullet$ For option (D) : The displacement of midpoint is

${L \over 2} - {L \over 2}\cos \theta = {L \over 2}(1 - \cos \theta )$

That is,

Displacement $\propto$ (1 $-$ cos$\theta$)

Hence, option (D) is correct.

We discuss the options as follows:

$\bullet$ For option (B) : From law of conservation of momentum, we have

mv = MV ...... (1)

where v is the velocity of point mass m and V is velocity of block of mass M.

Now, by using law of conservation of energy, after the point mass is released, we have the following:

Loss in P.E. of mass m = Gain in K.E. of both mass M and the mass m.

$ \Rightarrow mgh = {1 \over 2}m{v^2} + {1 \over 2}M{V^2}$

Substituting $V = {{mu} \over M}$ [from Eq. (1)], we get

$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{\left( {{{mv} \over M}} \right)^2}$

$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{{{m^2}{v^2}} \over {{M^2}}}$

$mgR = {1 \over 2}m{v^2} + {1 \over 2}{{{m^2}{v^2}} \over M}$

$mgR = {1 \over 2}m{v^2}\left( {1 + {m \over M}} \right)$

$gR = {1 \over 2}{v^2}\left( {1 + {m \over M}} \right)$

Rearranging this equation, we get

$2gR = {v^2}\left( {1 + {m \over M}} \right)$

${v^2} = {{2gR} \over {\left( {1 + {m \over M}} \right)}}$

$ \Rightarrow v = \sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $

Hence, option (B) is correct.

$\bullet$ For option (C) : Now, from Eq. (1), we have

$V = {m \over M}v = {m \over M}\sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $

$V = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {1 + {m \over M}} \right)}}} = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {{{M + m} \over M}} \right)}}} $

$V = \sqrt {{{{m^2}2gR} \over {M(m + M)}}} \Rightarrow V = m\sqrt {{{2gR} \over {M(m + M)}}} $

Since, there is no external force acting on the system, the centre of mass does not change.

Now, if the change in position of mass m is $\Delta$x and the change in position of mass M is $\Delta$X, then

m$\Delta$x + M$\Delta$x = 0 (since centre of mass will not change)

Now, we know that the change in position of point mass m is

$\Delta$x = R $-$ x

and the change in position of block of mass M is

$\Delta$X = $-$x

Therefore,

m(R $-$ x) $-$ Mx = 0

m(R $-$ x) = Mx

mR $-$ mx = Mx

mR = Mx + mx

mR = (M + m)x

$\Rightarrow$ x = ${{mR} \over {M + m}}$

The change in position of block of mass M is

$\Delta$x = $-$x = ${{ - mR} \over {M + m}}$

Thus, x-component of the displacement of the centre of mass of the block is

$M = {{ - mR} \over {M + m}}$

Hence, option (C) is also correct.

The following figure depicts the rolling condition of a circular disc:



For the smaller disc, we have the velocity expressed as

v = a$\omega$

For the bigger disc, we have the velocity expressed as

v = (2a)$\omega$ = 2a$\omega$

Now, let us consider that both discs roll and also rotate about point O.



$\bullet$ The angular momentum of a rotating body is given by

$\overrightarrow L = \overrightarrow r \times \overrightarrow p $

About point O, the combined angular momentum of the discs is given by

L = lma$\omega$ + 2l(4m)(2a$\omega$)

where lma$\omega$ is due to the smaller disc and 2l(4m) (2a$\omega$) is due to the bigger disc.

Now, substituting the values in Eq. (1), we get

$L = m{a^2}\omega [\sqrt {24} + 16\sqrt {24} ]$

$ = m{a^2}\omega (34\sqrt 6 ) \approx 83m{a^2}\omega $

Hence, option (B) is incorrect.

$\bullet$ Now, let us consider that z-component of $\overrightarrow L $ :

$\overrightarrow L = L\cos \theta $

$ = 34\sqrt 6 m{a^2}\omega \left[ {{1 \over {\sqrt {{l^2} + {a^2}} }}} \right] = 34\sqrt 6 m{a^2}\omega \left( {{{\sqrt {24} } \over 6}} \right)$

= 81 ma²$\omega$

Hence, option (D) is incorrect.

$\bullet$ The centre of mass of the assembly about its centre of mass [i.e. the axis lavelled as (I) in the figure shown here] is

${L_{(I)}} = {{m{a^2}} \over 2}\omega + {{4m{{(2a)}^2}} \over 2}\omega = {{17m{a^2}} \over 2}\omega $

Hence, option (C) is correct.


$\bullet$ About z-axis, the centre of mass of the assembly rotates with an angular speed

${\omega _z} = \omega \sin \theta = \omega \left[ {{a \over {\sqrt {{l^2} + {a^2}} }}} \right] = {\omega \over 5}$

Hence, option (A) is correct.

$r = \alpha {t^3}\widehat i + \beta {t^2}\widehat j$

$v = {{dr} \over {dt}} = 3\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$

$a = {{{d^2}r} \over {d{t^2}}} = 6\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$

At t = 1 s,

(a) $v = 3 \times {{10} \over 3} \times 1\widehat i + 2 \times 5 \times 1\widehat j$

$ = (10\widehat i + 10\widehat j)$ m/s

(b) $\widehat L = \widehat r \times \widehat p$

$ = \left( {{{10} \over 3} \times 1\widehat i + 5 \times 1\widehat j} \right) \times 0.1(10\widehat i + 10\widehat j)$

$ = \left( { - {5 \over 3}\widehat k} \right)$ N-ms

(c) $F = ma$

$ = m \times \left( {6 \times {{10} \over 3} \times 1\widehat i + 2 \times 5\widehat j} \right) = (2\widehat i + \widehat j)N$

(d) $\tau = r \times F = \left( {{{10} \over 3}\widehat i + 5\widehat j} \right) \times (2\widehat i + \widehat j)$

$ = + {{10} \over 3}\widehat k + 10( - \widehat k) = \left( { - {{20} \over 3}\widehat k} \right)$ N-m

Ring has mass M and radius R. Initial angular speed of ring = $\omega$. Two point masses, each of mass are at rest at O. Initial angular momentum of ring and point masses system,

${L_i} = {I_R}\omega + {I_m}\omega + {I_m}\omega $

$ = M{R^2}\omega + 0 + 0 = M{R^2}\omega $

After some time, situation is changed as shown in the figure.

Angular speed of the system, $\omega ' = {8 \over 9}\omega $; $OA = {{3R} \over 5}$; OB = r = ?

Moment of inertia about O of point mass at A,

${I_A} = {M \over 8} \times {{9{R^2}} \over {25}}$

Moment of inertia about O of point mass at B, ${I_B} = {M \over 8}{r^2}$

Fina angular momentum of the system

${L_f} = M{R^2}\omega ' + {I_A}\omega ' + {I_B}\omega '$

$ = M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$

As there is no external torque acting on the system so its angular momentum will be conserved, L_i = L_f

$M{R^2}\omega = M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$

${R^2} = {{8{R^2}} \over 9} + {{{R^2}} \over {25}} + {{{r^2}} \over 9} \Rightarrow {{{r^2}} \over 9} = {{16} \over {225}}{R^2}$ $\therefore$ $r = {4 \over 5}R$

The outer ring rolls without slipping, meaning the point on the ring in contact with the ground (denoted as point C) is stationary, i.e., $ \vec{v}_{\mathrm{C}} = \overrightarrow{0} $. Therefore, the velocity of point O is :

$ \vec{v}_{\mathrm{O}} = \vec{v}_{\mathrm{C}} + \vec{\omega}_{\mathrm{o}} \times \vec{r}_{\mathrm{CO}} = \overrightarrow{0} + \omega \hat{\jmath} \times 3R \hat{k} = 3R\omega \hat{\imath} $

Point P is located on the inner disc, which has an angular velocity $ \vec{\omega}_{\mathrm{i}} = -\omega / 2 \hat{\jmath} $. The position vector from O to P is given by :

$ \vec{r}_{\mathrm{OP}} = R \cos 30^{\circ} \hat{\imath} + R \sin 30^{\circ} \hat{k} = \frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k} $

Thus, the velocity of point P is calculated as follows :

$ \begin{aligned} \vec{v}_{\mathrm{P}} & = \vec{v}_{\mathrm{O}} + \vec{\omega}_{\mathrm{i}} \times \vec{r}_{\mathrm{OP}} \\ & = 3R\omega \hat{\imath} + \left(-\frac{\omega}{2} \hat{\jmath}\right) \times \left(\frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k}\right) \\ & = \frac{11\omega R}{4} \hat{\imath} + \frac{\sqrt{3} \omega R}{4} \hat{k} . \end{aligned} $

Since Cylinder P has most of its mass concentrated near its surface, its moment of inertia (about the cylinder axis) is greater than that of Cylinder Q, i.e., $ I_P > I_Q $. The forces acting on the cylinder include its weight $ mg $, the normal reaction $ N $, and the frictional force $ f $.

In the case of rolling without slipping, we have :

$ v = \omega r $

$ a = \alpha r $

The torque about the center of mass is related to $\alpha$ by :

$ \tau = rf = I\alpha \quad (1) $

The force along the plane is related to $a$ by :

$ mg \sin \theta - f = ma \quad (2) $

By solving equations (1) and (2), we get :

$ a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}} \quad (3) $

From equation (3), it follows that $ a_P < a_Q $ (since $ I_P > I_Q $). Therefore, Cylinder P reaches the ground later, has a lower velocity, lower angular velocity ($\omega = \frac{v}{r}$), and lower translational kinetic energy.

For verification, the conservation of energy can be used :

$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $

This yields :

$ \frac{1}{2}mv_P^2 < \frac{1}{2}mv_Q^2 $

Thus :

$ v_P < v_Q, \ \omega_P < \omega_Q, \ a_P < a_Q $

And the time relationships :

$ t_P > t_Q $

We have

Restoring torque = Force of gravity on disc and rod which is same in both cases.

$\bullet$ Case A : The moment of inertia is

${I_A} = {{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}$

Therefore,

${\tau _A} = - {I_A}\omega _A^2\theta = - \left( {{{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}} \right)\omega _A^2\theta $

$\bullet$ Case B : The moment of inertia is

${I_B} = {{m{l^3}} \over 3} + M{L^2}$

Therefore,

${\tau _B} = - {I_B}\omega _B^2\theta = - \left( {{{m{l^3}} \over 3} + M{L^2}} \right)\omega _B^2\theta $

since ${\tau _A} = {\tau _B} \Rightarrow {\omega _A} < {\omega _B}$.

We have

$\overrightarrow {{V_C}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_B}} = \overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_A}} = 0$

Therefore,

$\overrightarrow {{V_C}} - \overrightarrow {{V_B}} = \overrightarrow {r\omega } = \overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_B}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_C}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_B}} - \overrightarrow {{V_C}} = - \overrightarrow {r\omega } $

For pure rolling, we have

$|\overrightarrow {{V_{CM}}} | = |\overrightarrow {r\omega } |$

Therefore,

$|\overrightarrow {{V_C}} - \overrightarrow {{V_A}} | = 2{V_{CM}}$

$ \Rightarrow 2|\overrightarrow {{V_B}} - \overrightarrow {{V_C}} | = 2r\omega = 2{V_{CM}}$

According to Newton's second law, we have

$\sum {{{\overrightarrow f }_{ext}} = {{d\overrightarrow p } \over {dt}}} $

If $\sum {{{\overrightarrow f }_{ext}} = 0} $, we find that the linear momentum $\overrightarrow p $ as constant.

$ \text { From the diagram, } $

We observe $f$ opposes translational motion and (option d) supports rotational motion.

$ \begin{aligned} m g \sin \theta-f & =m a \\ f \mathrm{R} & =\mathrm{I} \alpha=\frac{1}{2} \mathrm{MR}^2\left(\frac{a}{\mathrm{R}}\right)=\frac{1}{2} m a \mathrm{R} \end{aligned} $

$ \begin{array}{rlrl} m g \sin \theta-f & =2 f & (m a & =2 f) \\ m g \sin \theta & =3 f & {[a} & =r \alpha \\ \alpha & \left.=\frac{a}{r}\right] & \end{array} $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f=\frac{1}{3} m g \sin \theta \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f \propto \sin \theta \end{aligned} $

Thus, if we reduce $\theta, f$ will be decreased Option (c) is correct.