Question

A body is moving in a vertical circle such that the velocities of body at
different points are critical. The ratio of velocities of body at angular
displacements 60° ,120° from lowest point is​

Answer 1

Answer:

$\frac{\sqrt{2}}{1}$

Explanation:

We know that,

The height of the body from the ground at any angle θ from the vertical can be given by,

$Height,h=R-Rcos\theta$

where,

R is the Radius of the Circle.

So,

The height of the body at 60° is,

$h_{1}=R-rcos60\\h_{1}=R-\frac{R}{2}\\h_{1}=\frac{R}{2}$

and,

The height of the body at 120° is,

$h_{2}=R-rcos120\\h_{2}=R+\frac{R}{2}\\h_{2}=\frac{3R}{2}$

Now,

We also know that,

K.E. + P.E. = Constant

i.e.

$\frac{1}{2}mv^{2}+mgh=k$

So,

$\frac{1}{2}mv_{1}^{2}+mgh_{1}=\frac{1}{2}mv_{2}^{2}+mgh_{2}\\\frac{1}{2}v_{1}^{2}+gh_{1}=\frac{1}{2}v_{2}^{2}+gh_{2}\\v_{1}^{2}-v_{2}^{2}=2g(h_{2}-h_{1})\\v_{1}^{2}-v_{2}^{2}=2g(\frac{3R}{2}-\frac{R}{2})\\v_{1}^{2}-v_{2}^{2}=2gR$

If the body just makes it to the top then we can say that,

$\frac{1}{2}mv_{1}^{2}=mg(2R)\\v_{1}^{2}=4gR$

On putting this in the previous equation we get,

$v_{1}^{2}-v_{2}^{2}=2gR\\4gR-v_{2}^{2}=2gR\\v_{2}^{2}=2gR\\v_{2}=\sqrt{2gR}$

Therefore, the ratio of the velocities will be,

$\frac{v_{1}}{v_{2}}=\frac{2\sqrt{gR}}{\sqrt{2gR}}\\\frac{v_{1}}{v_{2}}=\frac{\sqrt{2}}{1}$