Circular Motion #1

Classical Mechanics Level 2

A stone of mass $2 \text{ kg}$ performs uniform circular motion with speed $3 \text{ m s}^{ -1 }$ and radius $5 \text{ m}$ . What is the magnitude of the net force (in Newtons) acting on the stone?

This problem is part of the set - Circular Motion Practice

The answer is 3.6.

1 solution

Pranshu Gaba
Nov 4, 2015

Since the stone is performing uniform circular motion, there is no tangential acceleration; there is only radial (or centripetal) acceleration. The magnitude of radial acceleration is given by $v^{2} / r$ where $v$ is the velocity of the stone and $r$ is the radius of the circle it is moving in.

Thus, the net force acting on the stone is

$\begin{aligned} F_{net} & = ma \\ & = m \cdot \frac{ v^{2} } { r } \\ & = \frac{ 2 \times 3^{2} }{ 5 } \\ & = \boxed{ 3.6 \text{ N} } \ _\square \end{aligned}$

It would have been better if u had mentioned that the motion was done in an anti gravity space.

Aditya Kumar - 5 years, 7 months ago

Could you explain why it would have been better? I don't see how to presence or absence of gravity affects the problem. I think the problem is valid even if there is gravity.

Pranshu Gaba - 5 years, 7 months ago

Just imagine that the stone is performing UCM in a horizontal direction. There will also be a component of weight.

Aditya Kumar - 5 years, 7 months ago

@Aditya Kumar – If the stone is performing UCM in horizontal plane, then the component of weight will be cancelled by the normal reaction by floor on the stone.

Pranshu Gaba - 5 years, 7 months ago

@Pranshu Gaba – Horizontal plane need not mean that the body is in contact with any horizontal plane! Vertical motion does not mean that the stone undergoes UCM on a wall or something like that! A plane is just a 2-Dimensional Are which may be present anywhere in the 3-Dimensional Space! So, mentioning Anti-Gravity would have been useful!

Anirudh Krishna - 5 years, 6 months ago

Please clarify this question. If the stone is performing a circular motion, the displacement is zero. So, velocity is zero. So, acceleration is zero. F = m×a =m×0 = 0.

Ashish Menon - 5 years, 5 months ago

Circular Motion #1

The answer is 3.6.

1 solution

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