/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A block of mass \(m\) is revolvi... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of mass \(m\) is revolving in a smooth horizontal plane with a constant speed \(v .\) If the radius of the circular path is \(R\), find the total contact force received by the block. (1) \(\frac{m v^{2}}{R}\) (2) \(m g\) (3) \(m \sqrt{\frac{v^{4}}{R^{2}}+4 g^{2}}\) (4) \(m \sqrt{\frac{v^{4}}{R^{2}}+g^{2}}\)

Short Answer

Expert verified
The total contact force is option (4): \( m \sqrt{\frac{v^4}{R^2} + g^2} \).

Step by step solution

01

Understanding the Forces

A block in circular motion on a horizontal plane experiences two main forces: the centrifugal force due to its circular path and gravitational force acting vertically downward. We need to calculate both in order to determine the total contact force.
02

Centripetal Force Calculation

The centripetal force required to keep the block moving in a circle is given by the formula \( F_c = \frac{m v^2}{R} \), where \( m \) is the mass, \( v \) is the velocity, and \( R \) is the radius of the circular path.
03

Gravitational Force Identification

The force of gravity acting on the block is simply the weight of the block, which can be calculated as \( F_g = m g \), where \( g \) is the acceleration due to gravity.
04

Contact Force as the Resultant Force

The total contact force is the resultant force from both the centripetal force directed horizontally and the gravitational force directed vertically downward. These forces are perpendicular, so we can find the total contact force using the Pythagorean theorem: \( F_{ ext{total}} = \sqrt{F_c^2 + F_g^2} \).
05

Plugging in the Values

Substitute the known values: \( F_c = \frac{m v^2}{R} \) and \( F_g = m g \). Hence, the total contact force becomes \( F_{ ext{total}} = \sqrt{\left(\frac{m v^2}{R}\right)^2 + (mg)^2} = m \sqrt{\frac{v^4}{R^2} + g^2} \).

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Key Concepts

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

Centripetal Force
When an object moves in a circular path, it constantly changes direction. This change in direction is due to a force acting towards the center of the circle, called the centripetal force. In simple terms, this force keeps the object from flying off tangentially. For example, if you're swinging a ball on a string, the tension in the string is the centripetal force keeping the ball in its circular path.

The formula for calculating centripetal force is given by \( F_c = \frac{m v^2}{R} \), where:
  • \( m \) = mass of the object
  • \( v \) = velocity of the object
  • \( R \) = radius of the circular path
This formula shows that the force increases with the mass of the object and the square of its velocity. Additionally, a smaller circular path requires a larger force to maintain the same speed.

It's important to remember that centripetal force is not a new type of force. It's simply the result of other forces, like tension, gravity, or friction, acting towards the center of a circle.
Gravitational Force
Gravitational force is the attraction between two masses. On Earth, it pulls objects toward the center of the planet. It's the reason we don't float away into space. The force of gravity can be calculated using the formula \( F_g = m g \), where:
  • \( m \) = mass of the object
  • \( g \) = acceleration due to gravity (approximately \(9.8 \text{ m/s}^2\) on Earth)
This force acts downward, always pointing towards the ground.

Even when an object, like the block in this exercise, is moving in a horizontal circle, gravitational force still acts vertically downward. It's vital to consider gravity when determining the total contact force acting on the block. This is because it's one of the main components we consider along with the centripetal force.

Gravitational force doesn’t change as the block moves. Its consistency allows us to use it as a reliable component in various calculations and understandings of physical interactions.
Circular Motion
Circular motion occurs when an object moves in a path along a circle. This type of motion is fascinating because it involves forces and accelerations that constantly change direction. Unlike linear motion where objects move in a straight line, circular motion requires a constant inward force to bend the object's path.

Although the speed remains constant, the continuous change in direction means there's always acceleration toward the center of the circle. This inward acceleration is crucial, as it requires a centripetal force to sustain it. If this inward force were removed, the object would move off in a straight line due to inertia.

In problems involving circular motion, like the revolving block in this exercise, understanding the interplay between centripetal and gravitational forces is key. The block experiences centripetal force in the direction of the circle's center and gravitational force downward.

The combination of these forces informs us of the total contact force, which is the hypotenuse of the right triangle formed by these perpendicular forces. Thus, grasping the basics of circular motion helps in comprehending how forces interact in various physical systems.

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Most popular questions from this chapter

Ball \(A\) of mass \(m\), after sliding from an inclined plane, strikes elastically another ball \(B\) of same mass at rest. Find the minimum height \(h\) so that ball \(B\) just completes the circular motion of the surface at \(C\). (All surfaces are smooth.) (1) \(h=\frac{5}{2} R\) (2) \(h=2 R\) (3) \(h=\frac{2}{5} R\) (4) \(h=3 R\)

A small ring \(P\) is threaded on a smooth wire bent in the form of a circle of radius \(a\) and center \(O .\) The wire is rotating with constant angular speed \(\omega\) about a vertical diameter \(X Y\), while the ring remains at rest relative to the wire at a distance \(a / 2\) from \(X Y\). Then \(\omega^{2}\) is equal to (1) \(\frac{2 g}{a}\) (2) \(\frac{g}{2 a}\) (3) \(\frac{2 g}{a \sqrt{3}}\) (4) \(\frac{g \sqrt{3}}{2 a}\)

A small block of mass \(m\) slides on a frictionless horizontal table. It is constrained to move inside a ring of radius \(l\) which is fixed to the table. At \(t=0\), the block has a tangential velocity \(v_{0}\). The coefficient of friction between the block and the ring is \(\mu\). The velocity of the block at time \(t\) is (I) \(\frac{v_{0}}{1+\frac{\mu v_{0} t}{l}}\) (2) \(\frac{v_{0}}{1+\frac{2 \mu v_{0} t}{l}}\) (3) \(\frac{v_{0}}{1-\left(\frac{\mu v_{0} t}{l}\right)}\) (4) \(\frac{v_{0}}{\left(\frac{\mu v_{0} t}{l}\right)}\)

A particle tied at one end of an inextensible string of length \(R\), whose another end is fixed. Particle is performing circular motion in vertical plane. Ratio of maximum to minimum speed is \(\lambda\). Choose the correct option(s). (1) Speed at the top most point is \(\sqrt{\left(\frac{4 \lambda^{2}}{\lambda^{2}-1}\right) g R}\) (2) Speed at the top most point is \(\sqrt{\left(\frac{4}{\lambda^{2}-1}\right) g R}\) (3) Ratio of maximum to minimum tension is \(\left(\frac{5 \lambda^{2}-1}{5-\lambda^{2}}\right)\) (4) Ratio of tension in string to weight when string is horizontal is \(2\left(\frac{\lambda^{2}-1}{\lambda^{2}+1}\right)\)

\(A\) and \(B\) are moving in 2 circular orbits with angular velocity \(2 \omega\) and \(\omega\) respectively. Their positions are as shown at \(t=0 .\) Find the time when they will meet for first time. (1) \(\frac{\pi}{2 \omega}\) (2) \(\frac{3 \pi}{2 \omega}\) (3) \(\frac{\pi}{\omega}\) (4) they will never meet

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