Answer to Question #157198 in Mechanics | Relativity for wilder

Question

One end of a light inextensible string is fixed at a point A and a particle of mass mkg is attached to the other end B. When the particle moves in a horizontal circle of radius r below A with constant speed v m/s , the string is inclined at an angle u to the downward vertical. Show that v^2 = rgtanu.

Answer 1

Solution:

ABC is the triangle of vector forces where :

AC = m*g

AB = = m*g*cos u 

CB = T*sin u = $\tfrac{mV^2}{r}$

($\tfrac{mg}{cos(u)})*sin(u)$ =$\tfrac{mV^2}{r}$

mass m cancels 

g*tan u = $\tfrac{V^2}{r}$

$V^2=r*g*tan(u)$