Question

A small body of mass m tied to a non-stretchable thread moves over a smooth horizontal plane. The other end of the thread is being drawn into a hole $O$ (figure shown above) with a constant velocity. Find the thread tension as a function of the distance $r$ between the body and the hole if at $r=r_0$ the angular velocity of the thread is equal to ${\omega }_o$.

Answer 1

Hint: To solve this problem we use the concept of conservation of angular momentum since the momentum of force $T$ is also zero at the point $O$. Therefore, we can say that the angular momentum of the particle m is conserved about the centre point $O$.

Complete step by step answer:
Forces, acting on the mass m are shown in the figure. As a vector $\text{N = mg}$, the net torque of these two forces about any fixed point must be equal to zero. Tension T, acting on the mass m is a central force, which is always directed towards the centre O. Hence the moment of force T is also zero about the point O and therefore the angular momentum of the particle m is conserved about O.

Let, the angular velocity of the particle be $\omega$, when the separation between hole and particle m is r, then from the conservation of angular momentum about the point $O$,
$\therefore \text{m(}{\omega }_o{\text{r}}_o){\text{r}}_o=\text{m(}\omega \text{r)r}$
Also $\omega ={\displaystyle \frac{{\omega }_o{{\text{r}}_o}^2}{{\text{r}}^2}}$
Now from the Newton’s second law of motion,
$\text{T = F = m}{\omega }^2\text{r}$
$\Rightarrow \text{F = }{\displaystyle \frac{\text{m}{{\omega }_o}^2{{\text{r}}_o}^4\text{r}}{{\text{r}}^4}}={\displaystyle \frac{\text{m}{{\omega }_o}^2{{\text{r}}_o}^4}{{\text{r}}^3}}$


$\therefore$ The thread tension = ${\displaystyle \frac{\text{m}{{\omega }_o}^2{{\text{r}}_o}^4}{{\text{r}}^3}}$

Note:
Law of conservation of angular momentum states that if no external torque acts on the object, then there is no angular momentum.
$\because \text{torque (}\tau \text{) = }{\displaystyle \frac{\text{dL}}{\text{dT}}}$
$L$ = angular momentum
Because the momentum of $\tau$ is zero.
$\Rightarrow \tau =0$
$\Rightarrow {\displaystyle \frac{d\overrightarrow{L}}{d\overrightarrow{t}}}=0$
$\Rightarrow \text{L = 0}$
So, the angular momentum is conserved.
$\therefore \text{m(}\omega \text{r)r = m(}{\omega }_o{\text{r}}_o){\text{r}}_o$