Question

A rod of length L is hinged at one end and it is rotated with a constant angular velocity in a horizontal plane. Let $T_1$ and $T_2$ be the tensions at the points L/4 and 3L/4 away from the hinged end.

• A. $T_1>T_2$
• B. $T_2>T_1$
• C. $T_1=T_2$

Answer 1

Answer: (A)

Tension $T_1$ is greater than tension $T_2$ ($T_1 > T_2$).

Answer 2

The tension at any point in a rotating rod is due to the centripetal force required to keep the segment of the rod beyond that point in circular motion.

Let the rod have a total mass $M$ and length $L$. Its linear mass density is $\lambda = \frac{M}{L}$.

Consider a small element of the rod of length $dr$ at a distance $r$ from the hinged end. The mass of this element is $dm = \lambda dr$. This element, rotating with angular velocity $\omega$, experiences a centripetal force $dF = dm \omega^2 r = (\lambda dr) \omega^2 r$.

The tension $T(r_0)$ at a distance $r_0$ from the hinged end is the sum of the centripetal forces acting on all the mass elements from $r_0$ to $L$. This can be found by integrating $dF$ from $r_0$ to $L$:

$T(r_0) = \int_{r_0}^{L} \lambda \omega^2 r dr$ $T(r_0) = \lambda \omega^2 \int_{r_0}^{L} r dr$ $T(r_0) = \lambda \omega^2 \left[ \frac{r^2}{2} \right]_{r_0}^{L}$ $T(r_0) = \frac{\lambda \omega^2}{2} (L^2 - r_0^2)$

Substitute $\lambda = \frac{M}{L}$:

$T(r_0) = \frac{M \omega^2}{2L} (L^2 - r_0^2)$

Now, we calculate the tensions $T_1$ and $T_2$:

1. Tension $T_1$ at $r_0 = L/4$:

   $T_1 = \frac{M \omega^2}{2L} \left( L^2 - \left(\frac{L}{4}\right)^2 \right)$ $T_1 = \frac{M \omega^2}{2L} \left( L^2 - \frac{L^2}{16} \right)$ $T_1 = \frac{M \omega^2}{2L} \left( \frac{16L^2 - L^2}{16} \right)$ $T_1 = \frac{M \omega^2}{2L} \left( \frac{15L^2}{16} \right)$ $T_1 = \frac{15 M \omega^2 L}{32}$

2. Tension $T_2$ at $r_0 = 3L/4$:

   $T_2 = \frac{M \omega^2}{2L} \left( L^2 - \left(\frac{3L}{4}\right)^2 \right)$ $T_2 = \frac{M \omega^2}{2L} \left( L^2 - \frac{9L^2}{16} \right)$ $T_2 = \frac{M \omega^2}{2L} \left( \frac{16L^2 - 9L^2}{16} \right)$ $T_2 = \frac{M \omega^2}{2L} \left( \frac{7L^2}{16} \right)$ $T_2 = \frac{7 M \omega^2 L}{32}$

Comparison:

Comparing $T_1 = \frac{15 M \omega^2 L}{32}$ and $T_2 = \frac{7 M \omega^2 L}{32}$:

Since $15 > 7$, it is clear that $T_1 > T_2$. This is expected because the tension is maximum at the hinge and decreases as we move towards the free end (where it is zero). The point $L/4$ is closer to the hinge than $3L/4$.