2
$\begingroup$

Consider a point mass on the end of a string of length $l$ near the surface of the Earth. Initially it is hanging straight down, but it has initial speed $v_0$, such that it will undergo a circular trajectory.

Initial mechanical energy is $$E_{m_0}=\frac{mv_0^2}{2}$$ Mechanical energy at some point along the trajectory is $$E_{m_1}=\frac{mv_1^2}{2}+mgl(1-\cos{\theta})$$

Where the second term is the potential energy and $l(1-\cos{\theta})$ is the height of the particle mass from the initial position. $$v_1=\sqrt{v_0^2-2gl(1-\cos{\theta})}$$

Applying Newton's 2nd law we can obtain an expression for the tension force $$mg\cos{\theta}-T=-m\frac{v_0^2-2gl(1-\cos{\theta})}{l}$$ $$\implies T=\frac{m}{l}(v_0^2+3gl\cos{\theta}-2gl)$$ Where I simply used the radial acceleration $a_R=\frac{v^2}{l}$.

At the top of the circle $\theta=\pi$ and thus $T(\pi)=\frac{m}{l}(v_0^2-5gl)$.

For the tension force to make sense physically, it must be greater than or equal to zero. $$T\geq0 \implies v_0 \geq \sqrt{5gl}$$

Now I ask the question: what must $v_0$ be for the speed at the top of the circular trajectory to be $0$?

$$v_1(\pi)=\sqrt{v_0^2-4gl}=0$$ $$\implies v_0=\sqrt{4gl}$$

But a $v_0$ of $\sqrt{4gl}$ means the tension force must be negative, which would mean that it would be a force pulling at the particle mass radially away from the center of the circle. This isn't possible in this problem.

Does this mean that no matter what, there is no trajectory that has zero speed at the top of the circle?

$\endgroup$
1
  • $\begingroup$ I have just had a glimpse, but it looks like your equation for forces has the centripetal component direction opposite. $mgh-(T+\frac{mv^2}{l})=0 \rightarrow mgh-T=\frac{mv^2}{l}$? $\endgroup$
    – joseph h
    Oct 3, 2021 at 5:33

1 Answer 1

1
$\begingroup$

Does this mean that no matter what, there is no trajectory that has zero speed at the top of the circle?

Yes, if the mass is suspended by a string. Another way to see this is to realise that if the mass was stationary at the top of the circle then the string would have to support the weight of the mass so $T(\pi)=-mg$, which is impossible for a string.

If the mass is going slowly enough to stop at the top of the circle then the string will go slack before it reaches this point (in fact, when $\cos \theta = -\tfrac 2 3$).

If the mass is supported by a rod or a spring instead of a string then $T$ can be negative and the mass can now stop at the top of the circle.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.