Answer to Question #157511 in Mechanics | Relativity for Ulrich

Question

A particle of mass m lies on a smooth horizontal table and is attached by an inextensible string which passes through a smooth hole in the table, to a particle of mass 2m which hangs freely below the table. The particle of mass m describes a circle of radius 1m on the table with such uniform speed that the particle of mass 2m remains at rest. Calculate the uniform speed.

Answer 1

Explanations & Calculations

• As stated in the question all the contacts are smooth, thus friction neither supports nor restrict the motion of the two-mass system.
• Understand the situation that if the masses are left at rest, $\small 2m$ will drag the thread downwards & the thread will drag $\small m$ downwards through the hole. If there was some friction, it would have opposed this motion & kept them in rest.
• Since it is not, the mass $\small m$ needs to move in a circle on the table which can support & keep $\small 2m$ in place without falling down.
• Keeping a sufficient uniform speed, keeps $\small 2m$ stable whereas too low speeds cause $\small 2m$ go down & too much speed causes $\small 2m$ goes up & eventually breaking the stability.
• Centripetal force & the fictitious centrifugal force on m support this behavior.

• During this motion, mass $\small 2m$ is in vertical equilibrium, which leads to the possibility of writing, $\small \uparrow T= \downarrow2mg\cdots(1)$
• Since mass $\small m$ is in circular motion, Newton's second law can be applied on $\small m$ towards the center of the 1m circle.

$\qquad\qquad \begin{aligned} \small F&= \small ma\implies T=m\frac{v^2}{r}=m\frac{v^2}{1}=mv^2\cdots (2)\\ \small \end{aligned}$

• By (1) = (2),

$\qquad\qquad \begin{aligned} \small 2mg&= mv^2\\ \small v^2 &= \small 2g\cdots(\because\cancel{m})\\ \small\bold{v}&= \small \bold{\sqrt{2g}} \end{aligned}$

• This is the sufficient magnitude of the speed.