Question

A uniform rod AB length $\ell$ and mass m is suspended by two identical strings OA and OB from a fixed point O as shown in the figure. The rod is in the horizontal position and each string makes an angle $\theta$ with the rod. If the string OB is cut, choose the correct option (s) just after the string is cut: (Use $\theta$ = 30°)

• A. Tension in the string OA just after the cut is $\frac{2mg}{7}$
• B. Tension in the string OA just after the cut is $\frac{3mg}{7}$
• C. Angular acceleration of the rod just after the cut is $\frac{6g}{7\ell}$
• D. Angular acceleration of the rod just after the cut is $\frac{9g}{7\ell}$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The initial state of the rod is equilibrium. The rod AB of length $\ell$ and mass $m$ is suspended by two identical strings OA and OB from a fixed point O. The rod is horizontal, and each string makes an angle $\theta$ with the rod. Let T be the tension in each string.

In the initial equilibrium, the vertical forces balance the weight: $2T \sin \theta = mg$, so $T = \frac{mg}{2 \sin \theta}$. Given $\theta = 30^\circ$, $\sin 30^\circ = 1/2$. So, $T = \frac{mg}{2(1/2)} = mg$.

Just after the string OB is cut, the tension in OB becomes zero. The forces acting on the rod are the tension $T'$ in string OA and the weight $mg$ acting at the center of mass C (midpoint of AB). Let's consider the motion of the rod. Point A is constrained to move on a circle centered at O with radius equal to the length of string OA. Let $L$ be the length of string OA. In the initial configuration, consider the triangle formed by O, A, and the midpoint C of AB. AC = $\ell/2$. The angle OAC = $\theta$. Triangle OAC is a right-angled triangle with the right angle at C if OC is vertical and AB is horizontal. From the figure, the angle between the string and the rod is $\theta$. So, angle OAB = angle OBA = $\theta$. In triangle OAC, angle OAC = $\theta$, AC = $\ell/2$. $L = OA = AC / \cos \theta = \frac{\ell}{2 \cos \theta}$. Given $\theta = 30^\circ$, $\cos 30^\circ = \sqrt{3}/2$. So, $L = \frac{\ell}{2(\sqrt{3}/2)} = \frac{\ell}{\sqrt{3}}$.

Just after cutting the string OB, the rod starts to move. Let's apply Newton's laws and rotational dynamics. Let $\vec{a}_C$ be the acceleration of the center of mass C, and $\vec{\alpha}$ be the angular acceleration of the rod. The net force on the rod is $\vec{F}_{net} = \vec{T}' + m\vec{g} = m\vec{a}_C$. The net torque about the center of mass C is $\vec{\tau}_C = I_C \vec{\alpha}$. The moment of inertia of a uniform rod about its center is $I_C = \frac{1}{12} m \ell^2$.

Let's choose a coordinate system with origin at O, the fixed point. Let the initial position of O be (0, 0). Since the rod is horizontal and symmetric, the center of mass C is vertically below O. Let the y-axis be vertically downwards. Initially, O is at (0, 0). C is at $(0, h)$ where $h = OC$. In triangle OAC, $h = AC \tan \theta = (\ell/2) \tan \theta$. Given $\theta = 30^\circ$, $\tan 30^\circ = 1/\sqrt{3}$. So $h = \frac{\ell}{2\sqrt{3}}$. A is at $(-\ell/2, h)$ and B is at $(\ell/2, h)$. The string OA has length $L = \sqrt{(-\ell/2)^2 + h^2} = \sqrt{\ell^2/4 + \ell^2/12} = \sqrt{\frac{3\ell^2 + \ell^2}{12}} = \sqrt{\frac{4\ell^2}{12}} = \sqrt{\frac{\ell^2}{3}} = \frac{\ell}{\sqrt{3}}$.

Just after cutting OB, O is at (0, 0). Let the coordinates of A be $(x_A, y_A)$. The tension $\vec{T}'$ is along AO, so $\vec{T}' = -T' \frac{\vec{OA}}{|\vec{OA}|} = -T' \frac{(x_A, y_A)}{L}$. Let the coordinates of C be $(x_C, y_C)$. The weight is $m\vec{g} = (0, mg)$. $\vec{F}_{net} = \vec{T}' + m\vec{g} = m\vec{a}_C$. $-T' \frac{x_A}{L} \hat{i} + (-T' \frac{y_A}{L} + mg) \hat{j} = m(a_{Cx} \hat{i} + a_{Cy} \hat{j})$.

The torque about C is $\vec{\tau}_C = \vec{r}_{CA} \times \vec{T}'$. $\vec{r}_{CA} = \vec{A} - \vec{C} = (x_A - x_C, y_A - y_C)$. $\vec{\tau}_C = I_C \vec{\alpha}$.

Let's consider the acceleration constraints. Point A is connected to O by a string, so its acceleration $\vec{a}_A$ must have no component along the string OA. Let $\hat{u}_{OA}$ be the unit vector along OA. $\vec{a}_A \cdot \hat{u}_{OA} = 0$. $\hat{u}_{OA} = \frac{\vec{A} - \vec{O}}{|\vec{A} - \vec{O}|} = \frac{(x_A, y_A)}{L}$. $\vec{a}_A = \vec{a}_C + \vec{\alpha} \times \vec{r}_{CA}$. Initially, the rod is horizontal. Let the angle it makes with the horizontal be $\phi$. $\phi=0$ initially. Let the angular velocity be $\omega$. $\omega=0$ initially. Let the initial direction of AB be along the x-axis. C is at $(x_C, y_C)$. A is at $(x_C - \ell/2 \cos \phi, y_C - \ell/2 \sin \phi)$. Just after cutting, $\phi=0$ and $\omega=0$. Let the acceleration of C be $(a_{Cx}, a_{Cy})$ and angular acceleration be $\alpha$. $\vec{a}_A = \vec{a}_C + \vec{\alpha} \times \vec{r}_{CA}$. $\vec{r}_{CA} = (-\ell/2, 0)$ just after cutting, assuming the rod is along the x-axis through C. $\vec{\alpha} = (0, 0, \alpha)$ if rotation is in xy plane. $\vec{\alpha} \times \vec{r}_{CA} = (0, 0, \alpha) \times (-\ell/2, 0, 0) = (0, -\alpha (-\ell/2), 0) = (0, \alpha \ell/2, 0)$. $\vec{a}_A = (a_{Cx}, a_{Cy}) + (0, \alpha \ell/2) = (a_{Cx}, a_{Cy} + \alpha \ell/2)$.

The tension $\vec{T}'$ is along AO. The initial position of A relative to O is $(-\ell/2, h)$. The unit vector along OA is $\hat{u}_{AO} = \frac{(-\ell/2, h)}{L}$. The tension $\vec{T}'$ is along AO, so $\vec{T}' = T' \hat{u}_{AO} = T' \frac{(-\ell/2, h)}{L}$. $\vec{F}_{net} = \vec{T}' + m\vec{g} = T' \frac{(-\ell/2, h)}{L} + (0, mg) = m(a_{Cx}, a_{Cy})$. $a_{Cx} = -\frac{T' \ell}{2mL}$ and $a_{Cy} = \frac{T' h}{mL} + g$.

The torque about C is $\vec{\tau}_C = \vec{r}_{CA} \times \vec{T}'$. $\vec{r}_{CA} = (-\ell/2, 0)$. $\vec{T}' = T' \frac{(-\ell/2, h)}{L}$. $\vec{\tau}_C = (-\ell/2, 0, 0) \times \frac{T'}{L} (-\ell/2, h, 0) = (0, 0, (-\ell/2) \frac{T' h}{L} - 0) = (0, 0, -\frac{T' \ell h}{2L})$. $\tau_C = -\frac{T' \ell h}{2L}$. $I_C \alpha = \tau_C$. $\frac{1}{12} m \ell^2 \alpha = -\frac{T' \ell h}{2L}$. $\alpha = -\frac{6 T' h}{mL L}$. The negative sign indicates clockwise angular acceleration if $\alpha$ is taken as positive for counterclockwise. Let's assume positive $\alpha$ for clockwise. $\alpha = \frac{6 T' h}{mL L}$.

Constraint: $\vec{a}_A \cdot \hat{u}_{OA} = 0$. $\hat{u}_{OA} = \frac{(-\ell/2, h)}{L}$. $\vec{a}_A = (a_{Cx}, a_{Cy} + \alpha \ell/2)$. $(a_{Cx} \hat{i} + (a_{Cy} + \alpha \ell/2) \hat{j}) \cdot (-\ell/2 \hat{i} + h \hat{j}) = 0$. $a_{Cx} (-\ell/2) + (a_{Cy} + \alpha \ell/2) h = 0$. $-\frac{T' \ell}{2mL} (-\ell/2) + (\frac{T' h}{mL} + g + \alpha \ell/2) h = 0$. $\frac{T' \ell^2}{4mL} + \frac{T' h^2}{mL} + gh + \frac{\alpha \ell h}{2} = 0$. Substitute $\alpha = \frac{6 T' h}{mL L}$. $\frac{T' \ell^2}{4mL} + \frac{T' h^2}{mL} + gh + \frac{1}{2} \ell h \frac{6 T' h}{mL L} = 0$. Multiply by $mL$: $\frac{T' \ell^2}{4} + T' h^2 + mghL + 3 T' \ell h^2 / L = 0$. This seems wrong.

Let's consider the constraint in terms of acceleration components along and perpendicular to OA. $\vec{a}_A = \vec{a}_C + \vec{\alpha} \times \vec{r}_{CA}$. Let's use a coordinate system with origin at A. x-axis along AB, y-axis perpendicular to AB. $\vec{r}_{AC} = (\ell/2, 0)$. $\vec{C} = \vec{A} + \vec{r}_{AC}$. $\vec{a}_C = \vec{a}_A + \vec{\alpha} \times \vec{r}_{AC} - \omega^2 \vec{r}_{AC}$. Initially $\omega=0$. $\vec{a}_C = \vec{a}_A + \vec{\alpha} \times \vec{r}_{AC}$. Let $\vec{a}_A = a_{Ax} \hat{i} + a_{Ay} \hat{j}$ in this frame. $\vec{\alpha} = (0, 0, \alpha)$. $\vec{r}_{AC} = (\ell/2, 0, 0)$. $\vec{\alpha} \times \vec{r}_{AC} = (0, 0, \alpha) \times (\ell/2, 0, 0) = (0, \alpha \ell/2, 0)$. $\vec{a}_C = (a_{Ax}, a_{Ay} + \alpha \ell/2)$.

Forces on the rod: $\vec{T}'$ along AO, $\vec{mg}$ at C. $\vec{T}'$ makes angle $\theta$ with AB. Components are $(-T' \cos \theta, T' \sin \theta)$. $\vec{mg}$ acts vertically downwards. Let's find its components in this frame. The rod is horizontal, so gravity is along $-\hat{j}$. $\vec{mg} = (0, -mg)$. $\vec{F}_{net} = (-T' \cos \theta, T' \sin \theta - mg) = m(a_{Cx}, a_{Cy})$. $a_{Cx} = -\frac{T' \cos \theta}{m}$, $a_{Cy} = \frac{T' \sin \theta - mg}{m}$.

Torque about C: $\vec{\tau}_C = \vec{r}_{CA} \times \vec{T}' + \vec{r}_{CC} \times m\vec{g}$. $\vec{r}_{CC} = 0$. $\vec{r}_{CA} = (-\ell/2, 0)$. $\vec{T}' = (-T' \cos \theta, T' \sin \theta)$. $\vec{\tau}_C = (-\ell/2, 0, 0) \times (-T' \cos \theta, T' \sin \theta, 0) = (0, 0, (-\ell/2) T' \sin \theta - 0) = -\frac{T' \ell}{2} \sin \theta$. $I_C \alpha = \tau_C$. $\frac{1}{12} m \ell^2 \alpha = -\frac{T' \ell}{2} \sin \theta$. $\alpha = -\frac{6 T' \sin \theta}{m \ell}$. Let's assume positive $\alpha$ for clockwise rotation. $\alpha = \frac{6 T' \sin \theta}{m \ell}$.

Constraint on the motion of A. A is connected to O by a string. The acceleration of A perpendicular to OA is $a_{A \perp OA} = L \alpha_{OA}$, where $\alpha_{OA}$ is the angular acceleration of the string OA about O. The acceleration of A has no component along OA. Let the direction of OA be $\hat{u}_{OA}$. $\vec{a}_A \cdot \hat{u}_{OA} = 0$. Initially, the angle between AB and OA is $\theta$. Let's consider the angle of OA with the vertical. In triangle OAC, OC is vertical, AC is horizontal. Angle OAC is $\theta$. Angle OCA = $90^\circ$. Angle AOC = $90^\circ - \theta$. The angle of OA with the vertical is $90^\circ - \theta$. The angle with the horizontal is $\theta$. Let's use an inertial frame with y-axis vertical upwards and x-axis horizontal. Initial position of A relative to O is $(L \cos \theta, -L \sin \theta)$ if O is at (0, 0) and AB is horizontal below O. Let's put O at (0, 0). Initial position of A is $(x_A, y_A)$. String OA makes angle $\theta$ with the horizontal rod. The rod is horizontal. Let the rod be at height $y$ below O. $y = L \sin \theta$. Let the horizontal distance from O to C be 0. Then A is at $(-L \cos \theta, -L \sin \theta)$ and B is at $(L \cos \theta, -L \sin \theta)$. The length of the rod is $2 L \cos \theta = \ell$. So $L = \ell / (2 \cos \theta)$. This matches our previous result. Initial position of A relative to O is $(-\ell/2, -\frac{\ell}{2} \tan \theta)$. Let O be the origin (0, 0). Initial position of A is $(x_A, y_A) = (-\ell/2, -\frac{\ell}{2} \tan \theta)$. The tension $\vec{T}'$ is along OA. $\vec{T}' = -T' \hat{u}_{OA} = -T' \frac{(x_A, y_A)}{L} = -T' \frac{(-\ell/2, -\ell/2 \tan \theta)}{\ell/(2 \cos \theta)} = -T' \frac{(-\ell/2, -\ell/2 \sin \theta / \cos \theta)}{\ell/(2 \cos \theta)} = -T' \frac{(-\ell/2 \cos \theta, -\ell/2 \sin \theta)}{\ell/2} = -T' (-\cos \theta, -\sin \theta) = T' (\cos \theta, \sin \theta)$. This means the tension makes an angle $\theta$ with the horizontal, pointing towards O. This is consistent with the figure if the rod is below O.

Forces on the rod: $\vec{T}' = (T' \cos \theta, T' \sin \theta)$ at A, $\vec{mg} = (0, -mg)$ at C. Let the acceleration of C be $(a_{Cx}, a_{Cy})$ and angular acceleration be $\alpha$. $\vec{F}_{net} = \vec{T}' + m\vec{g} = (T' \cos \theta, T' \sin \theta - mg) = m\vec{a}_C$. $a_{Cx} = \frac{T' \cos \theta}{m}$, $a_{Cy} = \frac{T' \sin \theta - mg}{m}$.

Torque about C: $\vec{\tau}_C = \vec{r}_{CA} \times \vec{T}'$. $\vec{r}_{CA} = \vec{A} - \vec{C}$. Initial position of C is $(0, -\ell/2 \tan \theta)$. Initial position of A is $(-\ell/2, -\ell/2 \tan \theta)$. $\vec{r}_{CA} = (-\ell/2, 0)$. $\vec{\tau}_C = (-\ell/2, 0, 0) \times (T' \cos \theta, T' \sin \theta, 0) = (0, 0, -\ell/2 T' \sin \theta)$. $I_C \alpha = \tau_C$. $\frac{1}{12} m \ell^2 \alpha = -\frac{T' \ell}{2} \sin \theta$. $\alpha = -\frac{6 T' \sin \theta}{m \ell}$. Let's take positive $\alpha$ for clockwise rotation. $\alpha = \frac{6 T' \sin \theta}{m \ell}$.

Constraint: Acceleration of A has no component along OA. Initial unit vector along OA is $\hat{u}_{OA} = \frac{\vec{A} - \vec{O}}{|\vec{A} - \vec{O}|} = \frac{(-\ell/2, -\ell/2 \tan \theta)}{L} = \frac{(-\ell/2, -\ell/2 \sin \theta/\cos \theta)}{\ell/(2 \cos \theta)} = \frac{(-\ell/2 \cos \theta, -\ell/2 \sin \theta)}{\ell/2} = (-\cos \theta, -\sin \theta)$. $\vec{a}_A = \vec{a}_C + \vec{\alpha} \times \vec{r}_{CA}$. $\vec{a}_C = (a_{Cx}, a_{Cy})$. $\vec{\alpha} = (0, 0, \alpha)$. $\vec{r}_{CA} = (-\ell/2, 0, 0)$. $\vec{\alpha} \times \vec{r}_{CA} = (0, 0, \alpha) \times (-\ell/2, 0, 0) = (0, \alpha \ell/2, 0)$. $\vec{a}_A = (a_{Cx}, a_{Cy} + \alpha \ell/2)$. $\vec{a}_A \cdot \hat{u}_{OA} = (a_{Cx} \hat{i} + (a_{Cy} + \alpha \ell/2) \hat{j}) \cdot (-\cos \theta \hat{i} - \sin \theta \hat{j}) = 0$. $-a_{Cx} \cos \theta - (a_{Cy} + \alpha \ell/2) \sin \theta = 0$. $-(\frac{T' \cos \theta}{m}) \cos \theta - (\frac{T' \sin \theta - mg}{m} + \frac{1}{2} \ell \frac{6 T' \sin \theta}{m \ell}) \sin \theta = 0$. $-\frac{T' \cos^2 \theta}{m} - (\frac{T' \sin \theta - mg}{m} + \frac{3 T' \sin \theta}{m}) \sin \theta = 0$. $-\frac{T' \cos^2 \theta}{m} - (\frac{4 T' \sin \theta - mg}{m}) \sin \theta = 0$. Multiply by m: $-T' \cos^2 \theta - (4 T' \sin \theta - mg) \sin \theta = 0$. $-T' \cos^2 \theta - 4 T' \sin^2 \theta + mg \sin \theta = 0$. $T' (\cos^2 \theta + 4 \sin^2 \theta) = mg \sin \theta$. $T' (\cos^2 \theta + 4 (1 - \cos^2 \theta)) = mg \sin \theta$. $T' (\cos^2 \theta + 4 - 4 \cos^2 \theta) = mg \sin \theta$. $T' (4 - 3 \cos^2 \theta) = mg \sin \theta$. $T' = \frac{mg \sin \theta}{4 - 3 \cos^2 \theta}$. Given $\theta = 30^\circ$, $\sin \theta = 1/2$, $\cos \theta = \sqrt{3}/2$, $\cos^2 \theta = 3/4$. $T' = \frac{mg (1/2)}{4 - 3 (3/4)} = \frac{mg/2}{4 - 9/4} = \frac{mg/2}{(16-9)/4} = \frac{mg/2}{7/4} = \frac{mg}{2} \times \frac{4}{7} = \frac{2mg}{7}$. So, the tension in the string OA just after the cut is $\frac{2mg}{7}$. Option (A) is correct.

Now let's find the angular acceleration $\alpha = \frac{6 T' \sin \theta}{m \ell}$. $\alpha = \frac{6 (\frac{2mg}{7}) (1/2)}{m \ell} = \frac{6 \frac{mg}{7}}{m \ell} = \frac{6g}{7\ell}$. So, the angular acceleration of the rod just after the cut is $\frac{6g}{7\ell}$. Option (C) is correct.

Therefore, the correct options are (A) and (C).