Rotational Motion

Using parallel axis theorem, moment of inertia about the given axis in the figure below will be

$\begin{aligned} & I_1=\frac{2}{5} M R^2+M(2 R)^2 \\\\ & I_1=\frac{22}{5} M R^2\end{aligned}$

Considering both spheres at equal distance from the axis, moment of inertia due to both spheres about this axis will be

$ 2 I_1=2 \times 22 / 5 M R^2 $ = ${{44} \over 5}$MR²

Moment of inertia of rod

Here, given that L = 2R

$\therefore I_2=\frac{1}{12} M(2 R)^2=\frac{1}{3} M R^2$

So, total moment of inertia of the system is

$ \begin{aligned} I & =2 I_1+I_2=2 \times \frac{22}{5} M R^2+\frac{1}{3} M R^2 \\\\ \Rightarrow I & =\left(\frac{44}{5}+\frac{1}{3}\right) M R^2=\frac{137}{15} M R^2 \end{aligned} $

Centre of mass after the collision will be at O. Therefore,

$2m{L \over 6} = m{L \over 3}$

Now, using conservation of angular momentum, we get

$2m{L \over 6} \times v + m{L \over 3} \times 2v = I\omega $

Therefore,

$2m{L \over 6} \times v + m{L \over 3} \times 2v = \left[ {8m{{{L^2}} \over {12}} + 2m{{\left( {{L \over 6}} \right)}^2} + m{{\left( {{L \over 3}} \right)}^2}} \right]\omega $

$ \Rightarrow \left( {{L \over 6} + {L \over 3}} \right)2mv = \left[ {8m{{{L^2}} \over {12}} + {{2m{L^2}} \over {36}} + {{m{L^2}} \over 9}} \right]\omega $

$ \Rightarrow {{3L} \over 6} \times 2mV = {5 \over 6}m{L^2}\omega $

$ \Rightarrow Lmv = {5 \over 6}m{L^2}\omega \Rightarrow \omega = {{6Lmv} \over {5m{L^2}}} \Rightarrow \omega = {6 \over 5}{v \over L}$

When the end M is released and rod makes angle $\alpha$ with horizontal, the displacement of centre of mass is ${L \over 2}\sin \alpha $.

Now, we know $mg{L \over 2}\sin \alpha = {1 \over 2}I{\omega ^2}$

Here, $I = {{m{L^2}} \over 3}$; therefore,

$mg{L \over 2}\sin \alpha = {1 \over 2}{{m{L^2}} \over 3}{\omega ^2}$

$ \Rightarrow {\omega ^2} = {{3g\sin \alpha } \over L} \Rightarrow \omega = \sqrt {{{3g\sin \alpha } \over L}} \Rightarrow \omega \propto \sqrt {\sin \alpha } $

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.

Let n be the number of sides of a regular polygon. By symmetry, its centre of mass O will be equidistant from each vertex i.e., it lies at the centre of the circumscribed circle. Let r be the radius of circumscribed circle and h be the perpendicular distance of O from any side (see figure). The angle subtended by any side on the centre O is 2$\pi$/n and $\angle$PON = $\pi$/n.

When polygon rolls about the vertex P (without slipping or sliding), the point O moves in a circle of radius r centred at P. The point O reaches the maximum height (point O' in the figure) when PO' is perpendicular to PQ. Thus, the maximum increase in height of the locus of the centre of mass O is given by

$\Delta = r - h = {h \over {\cos (\pi /n)}} - h = h\left( {{1 \over {\cos (\pi /n)}} - 1} \right)$.

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.

For proper mixing, the concrete mixture should not stick to the top wall of the cylinder and it should fall down.

The maximum velocity at the top is v, then we have

${{m{v^2}} \over r} = mg$

$ \Rightarrow v = \sqrt {rg} $

Therefore, the maximum rotational speed of the drum is

$\omega = {v \over r} = \sqrt {{g \over r}} = \sqrt {{{9.8} \over {1.25}}} $ rad/s

$ = \sqrt {{{9.8} \over {1.25}}} \times {{60} \over {2\pi }}$ rpm = 26.74 $\approx$ 27 rpm

The angular momentum is

$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$

For the particle moving from $D \to A$, we have

$\overrightarrow L = m\left[ {\left( {{R \over {\sqrt 2 }}\widehat i + {R \over {\sqrt 2 }}\widehat j} \right) \times v( - \widehat j)} \right]$

$ = {{mvR} \over {\sqrt 2 }}( - \widehat k)$

For the particle moving from $A \to B$, we have

$\overrightarrow L = m\left[ {\left( {{R \over {\sqrt 2 }}\widehat i + {R \over {\sqrt 2 }}\widehat j} \right) \times v\widehat i} \right]$

$ = {{mvR} \over {\sqrt 2 }}( - \widehat k)$

For the particle moving from $C \to D$, we have

$\overrightarrow L = mv \times $ ($\bot$ distance) $\times$ $(\widehat k)$

$ = mv\left( {{R \over {\sqrt 2 }} + a} \right)(\widehat k)$

For the particle moving from $B \to C$, we get

$\overrightarrow L = mv \times $ ($\bot$ distance) $\times$ $(\widehat k)$

$ = mv\left( {{R \over {\sqrt 2 }} + a} \right)(\widehat k)$

The given particle of mass m experiences a centrifugal force:

F = ma = mr$\omega$²

$\Rightarrow$ a = r$\omega$²

$\Rightarrow$ ${{dv} \over {dt}} = r{\omega ^2}$

or, $\left( {{{dv} \over {dr}}} \right)\left( {{{dr} \over {dt}}} \right) = r{\omega ^2}$ ($\because$ ${{{dr} \over {dt}}}=v$)

$ \Rightarrow vdv = r{\omega ^2}dr$

$\int\limits_0^v {vdv = {\omega ^2}\int\limits_{R/2}^r {rdr} } $ ($\because$ at t = 0, $r = {R \over 2}$)

$ \Rightarrow {{{v^2}} \over 2} = {\omega ^2}\left( {{{{r^2}} \over 2}} \right)_{R/2}^r$

$ \Rightarrow v = \omega \sqrt {{r^2} - {{{R^2}} \over 4}} $

Now, $v = {{dr} \over {dt}}$

Therefore, ${{dr} \over {dt}} = \omega \sqrt {{r^2} - {{{R^2}} \over 4}} $

or, $\int\limits_{R/2}^r {{{dr} \over {\sqrt {{r^2} - {{{R^2}} \over 4}} }} = \omega \int\limits_0^t {dt} } $ ...... (1)

To solve this integral, we substitute $r = {R \over 2}\sec \theta $; therefore,

$dr = {R \over 2}\sec \theta \tan \theta d\theta $

$ \Rightarrow \int\limits_{{\pi \over 2}}^{{{\sec }^{ - 1}}(2r/R)} {{{(r/2)\sec \theta \tan \theta d\theta } \over {\sqrt {{{{R^2}} \over 4}{{\tan }^2}\theta } }} = \omega t} $

$ \Rightarrow \omega t = \int\limits_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)} {\sec \theta d\theta } $

$ = {\left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]^{{{\sec }^{ - 1}}(2r/R)}}$

$ = \left[ {\ln \left| {\sec \theta + \sqrt {{{\sec }^2}\theta - 1} } \right|} \right]_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)}$

$ \Rightarrow \omega t = \ln \left[ {{{2r} \over R} + {{\sqrt {4{r^2} - {R^2}} } \over R}} \right]$

Let ${{2r} \over R} = x$. Therefore,

${e^\omega } = x + \sqrt {{x^2} - 1} $

$ \Rightarrow ({e^{\omega t}} - x)^2 = {x^2} - 1$

$ \Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + x^2 = x^2 - 1$

$ \Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + 1 = 0$

$ \Rightarrow x = {{{e^{2\omega t}} + 1} \over {2{e^{\omega t}}}} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}$

On replacing x by ${{2r} \over R}$, we get

${{2r} \over R} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}$

or, $r = {R \over 4}[{e^{\omega t}} + {e^{ - \omega t}}]$

Hence, option (c) is correct.

The rotational quantity of velocity given by

${v_{rot}} = {{dr} \over {dt}} = {R \over 4}[\omega {e^{\omega t}} - \omega {e^{ - \omega t}}]$

${v_{rot}} = {{\omega R} \over 4}[{e^{\omega t}} - {e^{ - \omega t}}]$ ...... (1)

The force experienced by the block in the inertial frame is

${\overrightarrow F _{in}} = - m{\omega ^2}r\widehat i - mg\widehat k$ ..... (2)

Therefore, the force experienced by the block in the rotating frame is

${\overrightarrow F _{rot}} = {\overrightarrow F _{in}} + 2m({\overrightarrow v _{rot}} \times \omega \widehat k) + m(\omega \widehat k \times r\widehat i) \times \omega \widehat k$

${\overrightarrow F _{rot}} = ( - m{\omega ^2}r\widehat i - mg\widehat k) + 2m{{\omega R} \over r}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat i \times \omega \widehat k + m{\omega ^2}r\widehat i$

$ = - {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat j - mg\widehat k$

Therefore, the net reaction of the disc on the block is

${\overrightarrow F _{reaction}} - {\overrightarrow F _{rot}} = {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{\omega t}}]\widehat j + mg\widehat k$

Hence, option (B) is correct.

1. Determine the side length of the cube:

The cube with the maximum possible volume that can be cut from the sphere will have its diagonal equal to the diameter of the sphere. Let the side length of the cube be 'a'. Using Pythagoras in 3D, we have:

$a^2 + a^2 + a^2 = (2R)^2$

$3a^2 = 4R^2$

$a = \sqrt{\frac{4R^2}{3}} = \frac{2R}{\sqrt{3}}$

2. Calculate the mass of the cube:

The volume of the cube is $V = a^3 = \left(\frac{2R}{\sqrt{3}}\right)^3 = \frac{8R^3}{3\sqrt{3}}$

The density of the sphere (and hence the cube) is $\rho = \frac{M}{\frac{4}{3}\pi R^3}$

The mass of the cube is $m = \rho V = \frac{M}{\frac{4}{3}\pi R^3} \cdot \frac{8R^3}{3\sqrt{3}} = \frac{2M}{\sqrt{3}\pi}$

3. Find the moment of inertia of the cube:

The moment of inertia of a cube about an axis passing through its center and perpendicular to one of its faces is given by:

$I = \frac{1}{6}ma^2$

Substituting the values we found:

$I = \frac{1}{6} \cdot \frac{2M}{\sqrt{3}\pi} \cdot \left(\frac{2R}{\sqrt{3}}\right)^2$

$I = \frac{4MR^2}{9\sqrt{3}\pi}$

Therefore, the correct answer is Option A: ${{4M{R^2}} \over {9\sqrt {3\pi } }}$

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

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

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

As $\tau $ is perpendicular to the angular momentum $\overrightarrow L$ of the bob, so the direction of $L$ changes but magnitude remains same.