Question

(7) The total energy of a particle of mass m, revolved in a vertical circle of
radius r, is
$\frac{9}{2}$
mgr. The difference in its kinetic energies at the bottom
and top of the circle is
(a) mgr (b) 6mgr (c) 4mgr (d) 2mgr.​

Answer 1

Answer:

Answer is 2mgr

Here is why

Explanation:

Thanx for the question its great one

They asked for the difference between the kinetic energy so

KE

$kinetic \: energy= \frac{1}{2} m{v}^{2}$

Velocity of the particle at the bottom of the circle is given by (5gr)^0.5

Which nothing but square root of 5gr

And Velocity of the particle at the top of the circle is given by (gr)^0.5

Which nothing but square root of gr

Do you want to know how this velocities came then go through this link

Now kinetic energy =(m(v1-v2) ^2) /2

Which gives the answer as the 2mgr

Hope it helps

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