Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length l = \sqrt 24 a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is \omega . The angular momentum of the entire assembly about the point ‘O’ is \overrightarrow L (see the figure). Which of the following statement(s) is(are) true?

The center of mass of the assembly rotates about the z-axis with an angular speed of \omega \over 5, The magnitude of angular momentum of the assembly about its center of mass is 17ma^2\omega \over 2

The center of mass of the assembly rotates about the z-axis with an angular speed of \omega \over 5

The magnitude of angular momentum of center of mass of the assembly about the point O is 81\,ma^2\omega

The magnitude of angular momentum of the assembly about its center of mass is 17ma^2\omega \over 2

The magnitude of the z-component of \overrightarrow L is 55ma^2\omega

Rotational Motion PYQ — JEE Advanced 2016 · Shift 2 · Q40 with Solution

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Correct answer: A, C

Geometry of the assembly

Two discs are connected by a rigid rod through their centers.

Disc 1: mass $m$ , radius $a$
Disc 2: mass $4m$ , radius $2a$
Distance between centers: $l=\sqrt{24}\,a=2\sqrt{6}\,a$

Since the assembly rolls on a flat surface, the centers of the discs are at heights equal to their radii above the ground.

So if the smaller disc center is at height $a$ , the larger disc center is at height $2a$ . Hence the vertical separation of the centers is $\Delta z = 2a-a=a.$

If the rod makes angle $\theta$ with the horizontal plane, then $l^2=(\text{horizontal separation})^2+(\Delta z)^2.$ Thus, $(\text{horizontal separation})^2 = 24a^2-a^2=23a^2.$ So the horizontal projection of the rod has length $\sqrt{23}\,a.$

The assembly rotates about the vertical $z$ -axis through the instantaneous line joining the contact points with the ground.

Rolling constraint and angular speed of CM about the z-axis

The rod itself spins about its own axis with angular speed $\omega$ .

Let the assembly precess about the vertical $z$ -axis with angular speed $\Omega$ .

For pure rolling, each disc must satisfy $v_{C_i}=R_i\omega$ where $R_i$ is its radius and $v_{C_i}$ is the speed of its center due to precession about the vertical axis.

Let the perpendicular distances of the two centers from the $z$ -axis be $r_1$ and $r_2$ . Then $v_{C_1}=\Omega r_1=a\omega,$ $v_{C_2}=\Omega r_2=2a\omega.$ Hence, $r_2=2r_1.$

Also, the horizontal separation between the centers is $r_1+r_2=\sqrt{23}\,a.$ So,

\quad r_2=\frac{2\sqrt{23}}{3}a.$$ Then $$\Omega=\frac{a\omega}{r_1} = \frac{a\omega}{(\sqrt{23}/3)a}=\frac{3}{\sqrt{23}}\omega.$$ Now find the CM distance from the $z$-axis: $$R_{CM}=\frac{m r_1+4m r_2}{5m}=\frac{r_1+4r_2}{5}.$$ Substitute $r_2=2r_1$: $$R_{CM}=\frac{r_1+8r_1}{5}=\frac{9r_1}{5}=\frac{9\sqrt{23}}{15}a=\frac{3\sqrt{23}}{5}a.$$ Therefore speed of CM is $$v_{CM}=\Omega R_{CM}=\frac{3}{\sqrt{23}}\omega\cdot \frac{3\sqrt{23}}{5}a=\frac{9a\omega}{5}.$$ So the CM moves in a circle about the $z$-axis with angular speed $$\Omega=\frac{v_{CM}}{R_{CM}}=\frac{3}{\sqrt{23}}\omega.$$ This is **not** $\omega/5$. So **Option A is false**. --- 3. **Angular momentum of the center of mass about O** The total mass is $$M=5m.$$ Magnitude of orbital angular momentum of CM about point $O$ is $$L_{CM}=M R_{CM}^2\Omega.$$ Now, $$R_{CM}=\frac{3\sqrt{23}}{5}a, \quad \Omega=\frac{3}{\sqrt{23}}\omega.$$ Thus, $$L_{CM}=5m\left(\frac{3\sqrt{23}}{5}a\right)^2\left(\frac{3}{\sqrt{23}}\omega\right).$$ Compute: $$\left(\frac{3\sqrt{23}}{5}a\right)^2=\frac{207}{25}a^2.$$ Hence, $$L_{CM}=5m\cdot \frac{207}{25}a^2\cdot \frac{3}{\sqrt{23}}\omega =\frac{621}{5\sqrt{23}}ma^2\omega.$$ This is not equal to $81ma^2\omega$. So **Option B is false**. --- 4. **Angular momentum about the center of mass** We need the spin angular momentum of the assembly about its own CM. The rod axis is the common symmetry axis of the discs, and the assembly rotates about this axis with angular speed $\omega$. Moment of inertia of each thin disc about its own central symmetry axis: - Small disc: $$I_1=\frac{1}{2}ma^2$$ - Large disc: $$I_2=\frac{1}{2}(4m)(2a)^2=8ma^2$$ Since the rod is massless and passes through centers, the MOI of assembly about rod axis is $$I_{\text{rod-axis}}=I_1+I_2=\frac{1}{2}ma^2+8ma^2=\frac{17}{2}ma^2.$$ Therefore angular momentum about CM due to spin is $$L_{\text{about CM}}=I_{\text{rod-axis}}\omega=\frac{17}{2}ma^2\omega.$$ So **Option C is true**. --- 5. **$z$-component of total angular momentum about O** The total $z$-component has two parts: 1. orbital part of CM about $O$ 2. spin part about CM projected on $z$ First, orbital part: $$L_{CM,z}=M R_{CM}^2\Omega=\frac{621}{5\sqrt{23}}ma^2\omega.$$ Now the spin angular momentum is along the rod axis. Its magnitude is $$L_{spin}=\frac{17}{2}ma^2\omega.$$ The vertical component of rod direction is $$\cos\alpha=\frac{\Delta z}{l}=\frac{a}{\sqrt{24}a}=\frac{1}{2\sqrt{6}}.$$ So $$|L_{spin,z}|=L_{spin}\cos\alpha =\frac{17}{2}ma^2\omega\cdot \frac{1}{2\sqrt{6}} =\frac{17}{4\sqrt{6}}ma^2\omega.$$ Thus, $$|L_z|=L_{CM,z}+|L_{spin,z}|,$$ which is clearly not equal to $55ma^2\omega$. So **Option D is false**. --- 6. **Final conclusion** Only **Option C** is correct. --- 7. **Comparison with stored answer** Stored correct answer: **A, C** My derived answer: **C only** I therefore **disagree** with the stored answer. The likely issue is in Option A: the actual precession/angular speed about the vertical axis comes from the rolling constraint on both discs and is $$\Omega=\frac{3}{\sqrt{23}}\omega,$$ not $\omega/5$. Therefore A should be false.

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Correct answer: A, C

Geometry of the assembly

Two discs are connected by a rigid rod through their centers.

Disc 1: mass $m$ , radius $a$
Disc 2: mass $4m$ , radius $2a$
Distance between centers: $l=\sqrt{24}\,a=2\sqrt{6}\,a$

Since the assembly rolls on a flat surface, the centers of the discs are at heights equal to their radii above the ground.

So if the smaller disc center is at height $a$ , the larger disc center is at height $2a$ . Hence the vertical separation of the centers is $\Delta z = 2a-a=a.$

The assembly rotates about the vertical $z$ -axis through the instantaneous line joining the contact points with the ground.

Rolling constraint and angular speed of CM about the z-axis

The rod itself spins about its own axis with angular speed $\omega$ .

Let the assembly precess about the vertical $z$ -axis with angular speed $\Omega$ .

For pure rolling, each disc must satisfy $v_{C_i}=R_i\omega$ where $R_i$ is its radius and $v_{C_i}$ is the speed of its center due to precession about the vertical axis.

Let the perpendicular distances of the two centers from the $z$ -axis be $r_1$ and $r_2$ . Then $v_{C_1}=\Omega r_1=a\omega,$ $v_{C_2}=\Omega r_2=2a\omega.$ Hence, $r_2=2r_1.$

Also, the horizontal separation between the centers is $r_1+r_2=\sqrt{23}\,a.$ So,

\quad r_2=\frac{2\sqrt{23}}{3}a.$$ Then $$\Omega=\frac{a\omega}{r_1} = \frac{a\omega}{(\sqrt{23}/3)a}=\frac{3}{\sqrt{23}}\omega.$$ Now find the CM distance from the $z$-axis: $$R_{CM}=\frac{m r_1+4m r_2}{5m}=\frac{r_1+4r_2}{5}.$$ Substitute $r_2=2r_1$: $$R_{CM}=\frac{r_1+8r_1}{5}=\frac{9r_1}{5}=\frac{9\sqrt{23}}{15}a=\frac{3\sqrt{23}}{5}a.$$ Therefore speed of CM is $$v_{CM}=\Omega R_{CM}=\frac{3}{\sqrt{23}}\omega\cdot \frac{3\sqrt{23}}{5}a=\frac{9a\omega}{5}.$$ So the CM moves in a circle about the $z$-axis with angular speed $$\Omega=\frac{v_{CM}}{R_{CM}}=\frac{3}{\sqrt{23}}\omega.$$ This is **not** $\omega/5$. So **Option A is false**. --- 3. **Angular momentum of the center of mass about O** The total mass is $$M=5m.$$ Magnitude of orbital angular momentum of CM about point $O$ is $$L_{CM}=M R_{CM}^2\Omega.$$ Now, $$R_{CM}=\frac{3\sqrt{23}}{5}a, \quad \Omega=\frac{3}{\sqrt{23}}\omega.$$ Thus, $$L_{CM}=5m\left(\frac{3\sqrt{23}}{5}a\right)^2\left(\frac{3}{\sqrt{23}}\omega\right).$$ Compute: $$\left(\frac{3\sqrt{23}}{5}a\right)^2=\frac{207}{25}a^2.$$ Hence, $$L_{CM}=5m\cdot \frac{207}{25}a^2\cdot \frac{3}{\sqrt{23}}\omega =\frac{621}{5\sqrt{23}}ma^2\omega.$$ This is not equal to $81ma^2\omega$. So **Option B is false**. --- 4. **Angular momentum about the center of mass** We need the spin angular momentum of the assembly about its own CM. The rod axis is the common symmetry axis of the discs, and the assembly rotates about this axis with angular speed $\omega$. Moment of inertia of each thin disc about its own central symmetry axis: - Small disc: $$I_1=\frac{1}{2}ma^2$$ - Large disc: $$I_2=\frac{1}{2}(4m)(2a)^2=8ma^2$$ Since the rod is massless and passes through centers, the MOI of assembly about rod axis is $$I_{\text{rod-axis}}=I_1+I_2=\frac{1}{2}ma^2+8ma^2=\frac{17}{2}ma^2.$$ Therefore angular momentum about CM due to spin is $$L_{\text{about CM}}=I_{\text{rod-axis}}\omega=\frac{17}{2}ma^2\omega.$$ So **Option C is true**. --- 5. **$z$-component of total angular momentum about O** The total $z$-component has two parts: 1. orbital part of CM about $O$ 2. spin part about CM projected on $z$ First, orbital part: $$L_{CM,z}=M R_{CM}^2\Omega=\frac{621}{5\sqrt{23}}ma^2\omega.$$ Now the spin angular momentum is along the rod axis. Its magnitude is $$L_{spin}=\frac{17}{2}ma^2\omega.$$ The vertical component of rod direction is $$\cos\alpha=\frac{\Delta z}{l}=\frac{a}{\sqrt{24}a}=\frac{1}{2\sqrt{6}}.$$ So $$|L_{spin,z}|=L_{spin}\cos\alpha =\frac{17}{2}ma^2\omega\cdot \frac{1}{2\sqrt{6}} =\frac{17}{4\sqrt{6}}ma^2\omega.$$ Thus, $$|L_z|=L_{CM,z}+|L_{spin,z}|,$$ which is clearly not equal to $55ma^2\omega$. So **Option D is false**. --- 6. **Final conclusion** Only **Option C** is correct. --- 7. **Comparison with stored answer** Stored correct answer: **A, C** My derived answer: **C only** I therefore **disagree** with the stored answer. The likely issue is in Option A: the actual precession/angular speed about the vertical axis comes from the rolling constraint on both discs and is $$\Omega=\frac{3}{\sqrt{23}}\omega,$$ not $\omega/5$. Therefore A should be false.

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