Two masses m and m \over 2 are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k, at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is \tau = k \theta for angular displacement \theta . If the rod is rotated by \theta 0 and released, the tension in it when it passes through its mean position will be :

Center of Mass PYQ — JEE Main 2019 · 9 Jan · Shift 1 · Q56 with Solution

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Correct answer: C

Set up the geometry about the centre of mass

Let the distances of masses $m$ and $\dfrac m2$ from the suspension point (which is at the centre of mass of the system) be $r_1$ and $r_2$ respectively.

Since the total rod length is $l$ , $r_1+r_2=l$ and by centre of mass condition, $m r_1=\frac m2 r_2$ $2r_1=r_2$ So, $r_1+2r_1=l \Rightarrow 3r_1=l \Rightarrow r_1=\frac l3,\, r_2=\frac{2l}3$

Thus,

mass $m$ is at distance $\dfrac l3$
mass $\dfrac m2$ is at distance $\dfrac{2l}3$

Find the moment of inertia about the suspension axis

The rod is massless, so only the two masses contribute: $I=m\left(\frac l3\right)^2+\frac m2\left(\frac{2l}3\right)^2$ $I=\frac{ml^2}{9}+\frac m2\cdot \frac{4l^2}{9}$ $I=\frac{ml^2}{9}+\frac{2ml^2}{9}=\frac{ml^2}{3}$

Use energy conservation to find angular speed at mean position

Initially, the rod is twisted through angle $\theta_0$ and released from rest.

Initial torsional potential energy: $U=\frac12 k\theta_0^2$

When it passes through mean position, this becomes rotational kinetic energy: $\frac12 I\omega^2=\frac12 k\theta_0^2$ $I\omega^2=k\theta_0^2$

=\frac{k\theta_0^2}{ml^2/3} =\frac{3k\theta_0^2}{ml^2}$$ --- 4. **Find the tension in the rod at mean position** At the mean position, the rod has maximum angular speed $\omega$. Each mass needs centripetal force toward the suspension point. Consider the mass $m$ at distance $\dfrac l3$. The only force along the rod on this mass is the rod tension $T$. So, $$T=m\omega^2\left(\frac l3\right)$$ Substitute $\omega^2$: $$T=m\left(\frac{3k\theta_0^2}{ml^2}\right)\left(\frac l3\right)$$ $$T=\frac{k\theta_0^2}{l}$$ Check with the other mass $\dfrac m2$ at distance $\dfrac{2l}3$: $$T=\frac m2\,\omega^2\left(\frac{2l}3\right)=\frac{k\theta_0^2}{l}$$ which is consistent. --- 5. **Match with options** $$T=\frac{k\theta_0^2}{l}$$ So the correct option is: **C. $\dfrac{k\theta_0^2}{l}$** --- 6. **Compare with stored correct answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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View written solutionFree

Correct answer: C

Set up the geometry about the centre of mass

Let the distances of masses $m$ and $\dfrac m2$ from the suspension point (which is at the centre of mass of the system) be $r_1$ and $r_2$ respectively.

Since the total rod length is $l$ , $r_1+r_2=l$ and by centre of mass condition, $m r_1=\frac m2 r_2$ $2r_1=r_2$ So, $r_1+2r_1=l \Rightarrow 3r_1=l \Rightarrow r_1=\frac l3,\, r_2=\frac{2l}3$

Thus,

mass $m$ is at distance $\dfrac l3$
mass $\dfrac m2$ is at distance $\dfrac{2l}3$

Find the moment of inertia about the suspension axis

Use energy conservation to find angular speed at mean position

Initially, the rod is twisted through angle $\theta_0$ and released from rest.

Initial torsional potential energy: $U=\frac12 k\theta_0^2$

When it passes through mean position, this becomes rotational kinetic energy: $\frac12 I\omega^2=\frac12 k\theta_0^2$ $I\omega^2=k\theta_0^2$

=\frac{k\theta_0^2}{ml^2/3} =\frac{3k\theta_0^2}{ml^2}$$ --- 4. **Find the tension in the rod at mean position** At the mean position, the rod has maximum angular speed $\omega$. Each mass needs centripetal force toward the suspension point. Consider the mass $m$ at distance $\dfrac l3$. The only force along the rod on this mass is the rod tension $T$. So, $$T=m\omega^2\left(\frac l3\right)$$ Substitute $\omega^2$: $$T=m\left(\frac{3k\theta_0^2}{ml^2}\right)\left(\frac l3\right)$$ $$T=\frac{k\theta_0^2}{l}$$ Check with the other mass $\dfrac m2$ at distance $\dfrac{2l}3$: $$T=\frac m2\,\omega^2\left(\frac{2l}3\right)=\frac{k\theta_0^2}{l}$$ which is consistent. --- 5. **Match with options** $$T=\frac{k\theta_0^2}{l}$$ So the correct option is: **C. $\dfrac{k\theta_0^2}{l}$** --- 6. **Compare with stored correct answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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