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91Ó°ÊÓ

A particle of mass \(M\) moves with constant speed along a circul force \(F\). Its speed is (a) \(\sqrt{\frac{r F}{m}}\) (b) \(\sqrt{\frac{F}{r}}\) (c) \(\sqrt{F m r}\) (d) \(\sqrt{\frac{F}{m r}}\)

Short Answer

Expert verified
Option (a) is the correct answer.

Step by step solution

01

Write down the formula for centrifugal force

The formula for the centrifugal force F experienced by a particle moving in a circle of radius r with velocity v is given by \(F = M v^2 / r\).
02

Rearrange the formula to solve for v

We can rearrange the formula to find the velocity as \(v = \sqrt{F r / M}\)
03

Match the result with the given options

It can be seen that the expression for velocity, i.e., \(v = \sqrt{F r / M}\) is matching with option (a), where \(F = M v^2 / r\). So, the correct answer is option (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Motion
When we talk about circular motion, we're discussing objects that move in a path that takes the shape of a circle. The movement is continuous and keeps the object bound to a circular path due to a special type of force. The primary force at work here is called centripetal force, which means "center-seeking." It points toward the center of the circle, keeping the object from flying off.
A familiar example of circular motion is a car turning around a curve. As you might notice, the car stays on the curve thanks to the grip between the tires and the road. This gripping force acts as the centripetal force.
In essence, circular motion is about the consistent action of a force that directs an object's path into a circular shape, never letting it stray from the path.
Velocity Formula
Velocity in the context of circular motion is a bit different from straight line motion. In this case, we often focus on tangential velocity, the speed of an object moving along the edge of the circle. The formula for velocity in circular motion when centripetal force is known is derived from the centripetal force formula:
- Centripetal force formula: \( F = \frac{M v^2}{r} \)
- Solving for velocity, we get: \( v = \sqrt{\frac{F r}{M}} \)
This formula tells us that the velocity is directly related to both the force applied and the radius of the circle, and inversely related to the mass of the object. In simple terms, stronger force or a larger circle allows higher speed, while a more massive object requires more force to maintain the same speed.
Mass of Particle
The mass of a particle is a critical factor in circular motion. Mass is essentially the quantity of matter contained in an object. In the equation \( v = \sqrt{\frac{F r}{M}} \), the mass \( M \) directly influences the velocity. This is because a more massive object (higher \( M \)) requires a greater centripetal force to achieve the same speed as a lighter object.
Here's why mass matters:
  • Increased mass means more force is needed to keep an object in circular motion.
  • A lighter object speeds up more easily with the same force.

So, when considering circular motion, it's essential to understand that the mass affects how the force and velocity interplay to maintain the motion.
Radius of Circular Path
The radius of the circular path is the distance from the center of the circle to any point on its circumference. It plays a significant role in determining the motion of a particle traveling around a circle.
In the formula \( v = \sqrt{\frac{F r}{M}} \), the radius \( r \) is directly proportional to velocity. This means:
  • A larger radius allows for greater speeds if the force and mass remain constant.
  • This is because a larger circle provides a longer path to disperse rotational energy, resulting in higher velocities.

Therefore, the radius isn't just a measure of distance. It's a key factor influencing how quickly an object can move through a circular path under a given force.

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Most popular questions from this chapter

A mass of \(2 \mathrm{~kg}\) is tied to the end of a string of length \(1 \mathrm{~m}\). It is, tl. constant speed of \(5 \mathrm{~ms}^{-1}\). Given that \(g=10 \mathrm{~ms}^{-2}\). At which of the following locations of tension in the string will be \(70 \mathrm{~N}\) (a) At the top (b) At the bottom (c) When the string is horizontal (d) At none of the above locations

A smooth table is placed horizontally and an ideal spring of spring constant \(k=1000 \mathrm{~N} / \mathrm{m}\) and unextended length of \(0.5 \mathrm{~m}\) has one end fixed to its centre. The other end is attached to a mass of \(5 \mathrm{~kg}\) which is moving in a circle with constant speed \(20 \mathrm{~m} / \mathrm{s}\). Then the tension in the spring and the extension of this spring beyond its normal length are (a) \(500 \mathrm{~N}, 0.5 \mathrm{~m}\) (b) \(600 \mathrm{~N}, 0.6 \mathrm{~m}\) (c) \(700 \mathrm{~N}, 0.7 \mathrm{~m}\) (d) \(800 \mathrm{~N}, 0.8 \mathrm{~m}\)

A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 metres \(/ \mathrm{sec}\). The combined mass of the boy and the cycle is \(90 \mathrm{~kg}\). The angle that the cycle makes with the vertical so that it may not fall is \(\left(g=9.8 \mathrm{~m} / \mathrm{sec}^{2}\right)\) (a) \(60.25^{\circ}\) (b) \(63.90^{\circ}\) (c) \(26.12^{\circ}\) (d) \(30.00^{\circ}\)

A vehicle is moving with a velocity \(v\) on a curved road of width \(b\) and radius of curvature \(R\). For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is (a) \(\frac{v^{2} b}{R g}\) (b) \(\frac{r b}{R g}\) (c) \(\frac{v b^{2}}{R g}\) (d) \(\frac{v b}{R^{2} g}\)

A ball rolls off top of a staircase with a horizontal velocity \(u \mathrm{~m} / \mathrm{s}\). If the steps are \(h\) metre high and \(b\) mere wide, the ball will just hit the edge of nth step if \(n\) equals to (a) \(\frac{h u^{2}}{g b^{2}}\) (b) \(\frac{u^{2} 8}{g b^{2}}\) (c) \(\frac{2 h u^{2}}{g b^{2}}\) (d) \(\frac{2 u^{2} g}{h b^{2}}\)

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