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Question: A stone of mass \[m\] is tied to a string and is moved in a vertical circle of radius \(r\) making n...

A stone of mass mm is tied to a string and is moved in a vertical circle of radius rr making n revolution per minute. The total strength in the string when the stone is at its lowest point is
A. mgmg
B. m(g+πnr2)m\left( {g + \pi n{r^2}} \right)
C. m(g+nr)m\left( {g + nr} \right)
D. m(g+π2n2r900)m\left( {g + \dfrac{{{\pi ^2}{n^2}r}}{{900}}} \right)

Explanation

Solution

We consider motion non-uniform because of the gravitational force in a vertical circular motion.
The objects and velocity and tension gets varied in magnitude in different directions
When the axis of the circle is horizontal then the motion is called the vertical circular motion.

Complete step by step answer:
The speed of the object and the direction of the object is changing constantly hence the gravity evenly speeds up the object or slows down the speed
Uniform vertical circular motion describes that when an object is moving in a circular motion, then the plane of the circle becomes vertical, here the object’s speed does not change when it moves around the circle.
When the tension in the string reaches the lowermost point,
T=mg+mω2rT = mg + m{\omega ^2}r
Now substitute ω=2πn\omega = 2\pi n in the above equation,
T=m(g+4π2n2r)T = m\left( {g + 4{\pi ^2}{n^2}r} \right)
Here nn revolution per minute hence the above equation becomes,
T=m(g+4π2(n60)2r)T = m\left( {g + 4{\pi ^2}{{\left( {\dfrac{n}{{60}}} \right)}^2}r} \right)
Now simplify the above equation,
T=m(g+(π2n2r900)2)T = m\left( {g + {{\left( {\dfrac{{{\pi ^2}{n^2}r}}{{900}}} \right)}^2}} \right)

So, the correct answer is “Option D”.

Note:
Due to the combination of gravitational, centripetal force, and tension the uniform vertical circular motion is possible.
It occurs if an object tied to a string is given the small required kinetic energy or velocity which completes the circle without slack.
An example of the vertical circular motion is swimming buckets of water overhead.