Rotational Motion

The moment of inertia $ I_{\text{cm}} $ of an annular disc, with an inner radius $ a $, an outer radius $ b $, and a mass $ m $, about an axis perpendicular to the disc and passing through its center $ C $, is given by :

$ I_{\text{cm}} = \frac{1}{2} m \left( a^2 + b^2 \right) $,

where $ a^2 $ and $ b^2 $ are the squares of the inner and outer radii, respectively.


The linear impulse $ J_0 $ alters the linear momentum of the disc. Right after the impulse is applied, the linear momentum can be represented as :

$ m v_{\text{cm}} = J_0 $,

where $ m $ is the mass of the disc and $ v_{\text{cm}} $ is the velocity of the center of mass resulting from the impulse.

The angular impulse about the center of mass $ C $ of the disc is applied in a clockwise direction, with its magnitude being $ J_{0, \text{ang}} = h J_0 $. This impulse changes the angular momentum of the disc about point $ C $, expressed as :

$ I_{\text{cm}} \omega = J_{0, \text{ang}} = h J_0 $,

where $ I_{\text{cm}} $ is the moment of inertia of the disc about its center of mass, $ \omega $ is the angular velocity resulting from the angular impulse, $ J_0 $ is the linear impulse, and $ h $ is a factor relating the angular impulse to the linear impulse.

For the disc to roll without slipping, given that the friction coefficient $ \mu $ is non-zero, the condition is $ v_{\text{cm}} = \omega b $. By substituting the expressions for $ v_{\text{cm}} $ and $ \omega $ and solving, we obtain the following relationship for the minimum value of $ h $ (denoted as $ h_m $) :

$ h_m = \frac{I_{\text{cm}}}{mb} = \frac{a^2 + b^2}{2b} $,

where $ I_{\text{cm}} $ is the moment of inertia of the disc about its center of mass, $ m $ is the mass of the disc, $ a $ is the inner radius, and $ b $ is the outer radius of the disc.

By substituting $ a = 0 $ into the formula, we find that $ h_m = \frac{b}{2} $. This implies that a solid disc will roll without slipping if the impulse is applied midway along the upper half of the disc.

Similarly, substituting $ a = b $ yields $ h_m = b $. This indicates that a ring (an annular disc with equal inner and outer radii) will roll without slipping when the horizontal impulse is applied at the very top of the ring.

For the condition where $ h = h_m $, the initial angular velocity $ \omega $ is given by :

$ \omega = \frac{h_m J_0}{I_{\text{cm}}} = \frac{J_0}{mb} $.

This expression for the angular velocity is independent of the inner radius $ a $ of the disc.

When a wheel is placed on a smooth surface with no friction ($\mu = 0$) and an impulse is applied directly at its center ($h = 0$), there is no torque generated to initiate rotation. Consequently, under these conditions, if an impulse is delivered to the center of a wheel on a frictionless surface, the wheel will invariably slide and not roll.

F $-$ f = ma_c

fR = I_C$\alpha$

a_C $-$ $\alpha$R = 0

$F - {I_C}{\alpha \over R} = m{a_c}$

${a_C} = {F \over {{{{I_C}} \over {{R^2}}} + m}}$

$f = {{{I_C}\alpha } \over R} = {{{I_C}} \over {{R^2}}},{a_C} = {{{I_C}} \over {{R^2}}}{F \over {\left[ {{{{I_C}} \over {{R^2}}} + m} \right]}}$

$f = {F \over {\left( {m + {{{I_C}} \over {{R^2}}}} \right)}}$

Thin walled hollow cylinder,

I_C = mR²

${a_C} = {F \over {2m}}$

$fR = {I_C}\alpha = {{{I_C}{a_C}} \over R}$

$f = {{{I_C}{a_C}} \over {{R^2}}} \le \mu mg$

${a_C} \le {{\mu mg{R^2}} \over {{I_C}}}$

For solid cylinder, ${I_C} = {{m{R^2}} \over 2}$

${a_C} \le 2\mu g$

${({a_C})_{\max }} = 2\mu g$

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.