Question

The total energy in the string revolving in a vertical circle with a mass m, radius 'r' and
acceleration due to gravity 'g' at the lowest point is.
a)5/2 mgr
b)2/3 mgr
c) 6 mgr
d) 3 mgr​

Answer 1

Answer:

5/2mgr

Explanation:

Velocity at lowest point(v)=

$\sqrt{5gr}$

kinetic energy=

$\frac{1}2(m)(v) {}^{2}$

$= \frac{1}2(m)( \sqrt{5gr) {}^{2} } {}$

$= \frac{5}2(mgr){}$

Kinetic energy

Answer 2

Answer:

The correct answer is Option A : $\frac{5}{2}mgr$

Explanation:

Velocity at the lowest point is $(v)=\sqrt{5gr}$

We know that the formula for kinetic energy :

$\frac{1}{2}mv^2$

$=\frac{1}{2}m(\sqrt{5gr})^2$

$=\frac{1}{2}m(5gr)$

$=\frac{5}{2} mgr$

Energy of an object traveling in a vertical circle:

• The body's energy is divided into two categories: kinetic energy and potential energy.

• On the circular route, the lowest point has the most kinetic energy, while the highest point has the least.

• Energy is conserved in its entirety.