Question

A smooth semicircular wire track of radius R is fixed in a vertical plane (see $Fig.6.201$). One end of a massless spring of natural length (3/4) R is attached to the lowest point o of the wire track. A small ring of mass m, which can slide on the track, A small ring of mass m, which can slide on the track, is attached to the other end of the spring. The ring is held stationary of point P such that the spring makes an angle of ${60}^0$ with the vertical. The spring constant $k=\left(mg/R\right)$. Consider the instant when the ring is released. Determine the tangential acceleration of the ring and the normal reaction between ring and track.

Answer 1

1. The free-body diagram of the ring is shown in $Fig.6.202$. The forces acting on the ring are :
a. The weight mg acting vertically downwards
B. Normal force N by the wire track.
Normal force on the ring could be either radially outwards or radially inwards depending on whether the ring presses against the inner surface or outer surface of the track. To ascertain whether normal force is inwards or outwards, assume that, to begin with, it is inwards. Then from $\sum \overrightarrow{}=m\overrightarrow{a}$,find the value of normal force. If it is positive. it is inwards and if it is negative, it is outwards.
C. Force of the spring kx. In the given physical situation, the spring is extended, it will pull the ring. So the spring force kx is along the spring towards o.
Length of the spring in the position shown= R.

$CP=R$

$\angle COP=\angle OPC={60}^0$

$△COP=Rightangle△$

Change in length of the spring $=R-\left(\frac{3}{4}\right)R=\left(\frac{R}{4}\right)$
$kx=\left(\frac{mg}{R}\right)\left(\frac{R}{4}\right)=\frac{mg}{4}$
Now from $F_t={ma}_{1^{\prime }}$
$\left(\frac{mg}{4}\right)\cos {30}^0+mg\cos {30}^0={ma}_1\Rightarrow a_1=\frac{5\sqrt{3}}{8}g$