Question

A small block of mass $m$ is tied to a spring of spring constant $k$ and length $L$. The other end of spring is fixed at a particular point. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity $\omega$ about point, then tension in the spring is

Answer 1

$T = \frac{kmω²L}{k - mω²}$

Answer 2

The block moves in a circular path, requiring a centripetal force $F_c$. This force is provided by the tension $T$ in the spring.

1. Centripetal Force: The centripetal force required for circular motion is $F_c = mω²r$, where $m$ is the mass, $\omega$ is the angular velocity, and $r$ is the radius of the circular path.
2. Spring Tension: The tension in the spring is given by Hooke's Law, $T = kx$, where $k$ is the spring constant and $x$ is the extension of the spring.
3. Radius of Path: The radius of the circular path $r$ is the sum of the spring's natural length $L$ and its extension $x$. So, $r = L + x$.
4. Equating Forces: Since the spring tension provides the centripetal force, we have $T = F_c$. Substituting the expressions: $kx = mω²(L + x)$
5. Solving for Extension (x): $kx = mω²L + mω²x$ $kx - mω²x = mω²L$ $x(k - mω²) = mω²L$ $x = \frac{mω²L}{k - mω²}$ (This solution is valid only if $k - mω² > 0$, i.e., $k > mω²$.)
6. Calculating Tension (T): Substitute the expression for $x$ back into $T = kx$: $T = k \left( \frac{mω²L}{k - mω²} \right)$ $T = \frac{kmω²L}{k - mω²}$