Question

11.
A heavy particle hanging from a string of length /
is projected horizontally with speed Vgl. Find the
speed of the particle at the point where the tension
in the string equals weight of the particle.
(1) √391 (2) ▼zgl​

Answer 1

Explanation:

This is Circular Motion

By Centeroid Acceleration thoery

mg=mU^2/L

so root( gl)=u

But .The datas are too much

I think that.This should be Projected Vertically.(Gravity)

Then.

if the String Changed a Angle in This tension.

Using F=ma to the Center

mg-mgcos a=mv^2/r

in Gravity.so using Motion formulas

V^2=gl-(2gh)

h=l-lcos a

so. Now.

g-g cos a=gl-2gl+2gl cos a=(2 gl cos a-gl)/L

now 1-cos a=2cos a-1

now Cos a=2/3

So V^2=(2gL×2/3)-Gl=gl/3

So.V=root (gl/3)

now.V^2=