in this illustration we’ll analyze, the situation
of a rod supported by strings. here the figure shows a rod ay b length of l and mass m supported
by 2 strings which are of equal length and in extensible. and here we are required to
find tension in 1 string just after other string is cut. so, in, situation we can analyze
when 1 string is cut the situation would be like this, say, this is the rod. and this
is only in the support of. the string at end b and string at end ay is say snapped, so
due to this what happens. the weight of rod will cause, a rotational motion in rod, and
a tension in the other string which remaining is say t. then we can write. if acceleration
of, center of mass, of rod, just after. 1 string is cut. is ay, and, angular acceleration.
is alpha. then here we use, we can write 1 equation as, m g minus t is equal to m ay.
because center of mass is coming downward acceleration ay, and if it is rotating at
an angular acceleration alpha about end b. then, the torque of tension about b is zero
only due to the torque of m g it will rotate. so we can write m g multiplied by. l by 2.
is equal t, i alpha which is m l square by 3. multiplied by its angular acceleration
will be, ay upon, l by 2 in this case because, the acceleration of center of mass we can
write as l by 2 alpha. so, if this is our equation 1 and this is equation 2. we can
simplify that equations and here we can get, from. equation 2. here the value we are getting
for, ay is, 3 gee by, 4. now, from equation 1. we can write m g minus t is equal to m
multiplied by 3 gee by 4. and simplifying this is giving us the value of t. that is
m g minus 3 em g by, 4 that is equal to, m g by, 4 that is a result this problem.