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A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis with a uniform angular speed \(2 \mathrm{rad} / \mathrm{s}\). The radius of the disc is \(50 \mathrm{~cm}\). Find the minimum coefficient of friction between disc and coin so that the coin ?oes not slip \(\left(g=10 \mathrm{~ms}^{-2}\right)\). a. \(0.1\) b. \(0.2\) c. \(0.3\) d. \(0.4\)

Short Answer

Expert verified
The minimum coefficient of friction is 0.2 (option b).

Step by step solution

01

Understanding the Problem

We are given a rotating disc with a coin placed at its edge. The goal is to find the minimum coefficient of friction that prevents the coin from slipping off the disc. The disc has an angular speed of \(2 \, \mathrm{rad/s}\) and a radius of \(50 \, \mathrm{cm}\), which is \(0.5 \, \mathrm{m}\). The value of gravitational acceleration \(g\) is \(10 \, \mathrm{ms}^{-2}\).
02

Calculating Centripetal Force

The centripetal force \(F_c\) required to keep the coin in circular motion is given by the formula \(F_c = m \omega^2 r\), where \(m\) is the mass of the coin, \(\omega = 2 \, \mathrm{rad/s}\) is the angular speed, and \(r = 0.5 \, \mathrm{m}\) is the radius of the disc.
03

Finding Frictional Force Requirement

The frictional force \(f\) is what prevents the coin from slipping, and it must be equal to the centripetal force for the coin to remain in position: \(f = F_c = m \omega^2 r\). Since frictional force \(f\) is given by \(f = \mu mg\), where \(\mu\) is the coefficient of friction, we equate this to the centripetal force: \(\mu mg = m \omega^2 r\).
04

Solving for the Coefficient of Friction

By cancelling the mass \(m\) from both sides of \(\mu mg = m \omega^2 r\), we derive \(\mu = \frac{\omega^2 r}{g}\). Substitute \(\omega = 2 \, \mathrm{rad/s}\), \(r = 0.5 \, \mathrm{m}\), and \(g = 10 \, \mathrm{m/s^2}\) into the equation: \[ \mu = \frac{(2)^2 \times 0.5}{10} = \frac{2}{10} = 0.2 \]
05

Checking the Options

The minimum coefficient of friction is calculated to be 0.2. Checking against the given options, option b is \(0.2\). Thus, this is the correct answer, ensuring the coin does not slip.

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Key Concepts

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

Coefficient of Friction
The coefficient of friction is a measure of how much frictional force exists between two surfaces. It is a crucial factor in determining whether one object will slip over another, especially in scenarios involving motion.

In the context of a rotating disc with a coin at its edge, the coin tends to slip due to the rotational motion. The force that prevents this is the frictional force between the coin and disc. This frictional force can be calculated using the coefficient of friction, denoted as \( \mu \).
  • The equation for frictional force is given by \( f = \mu mg \), where \( m \) is mass, and \( g \) is gravitational acceleration.
  • If frictional force equals the centripetal force required for circular motion, the coin will not slip.
  • Thus, \( \mu = \frac{\omega^2 r}{g} \) was used to ensure equilibrium in this exercise.
By understanding this principle, we can see why the coefficient of friction must be at a certain minimum level to prevent slipping. This ensures the dynamic stability of systems operating under rotational motion.
Rotational Motion
Rotational motion involves objects moving around a central axis. This movement is characterized by angular speed and radius, which are essential in determining forces at play.

For the rotating disc scenario, rotational motion was enabling the coin to trace a circular path around the axis of the disc:
  • Angular speed \( \omega = 2 \, \mathrm{rad/s} \) is the rate at which the coin rotates around the axis.
  • Radius \( r = 0.5 \, \mathrm{m} \) is the distance from the axis to the coin's position.
  • The balance of forces in rotational motion ensures that the coin remains on its path without slipping.
Rotational dynamics require that all moving parts experience centripetal force that keeps them in a circular path. Without sufficient centripetal force, derived from factors like friction, the object would cease its circular motion and potentially fly off at a tangent to the circle. Grasping these basic principles is key to understanding motion in a rotational framework.
Centripetal Acceleration
Centripetal acceleration is the acceleration directed towards the center of a circular path. It is necessary for maintaining an object's circular motion.

In our exercise:
  • Centripetal acceleration \( a_c \) arises due to the change in direction of the velocity vector as the coin moves in the circle.
  • It ensures that the speed remains constant while the direction changes steadily.
  • Calculated using \( a_c = \omega^2 r \), where \( \omega \) is the angular speed and \( r \) the radius of the disc.
This acceleration is essential because it generates the centripetal force needed to maintain circular motion. The frictional force becomes critical here as it must match or exceed this centripetal force to prevent the coin from slipping. Understanding how centripetal acceleration works allows us to comprehend why objects need sufficient friction to stay on rotating discs or any circular tracks.

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Most popular questions from this chapter

A particle of small mass \(m\) is joined to a very heavy body by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is a. \(>>m g\) b. \(4 \mathrm{mg}\) c. \(2 \mathrm{mg}\) d. \(\overline{m g}\)

Friction force can be reduced to a great extent by a. Lubricating the two moving parts. b. Using ball bearings between two moving parts. c. Introducing a thin cushion of air maintained between two relatively moving surfaces. d. All of the above.

A \(60 \mathrm{~kg}\) man stands on a spring scale in a lift. At some instant, he finds that the scale reading has changed from \(60 \mathrm{~kg}\) to \(50 \mathrm{~kg}\) for a while and then comes back to original mark. What should be concluded? a. The lift was in constant motion upwards. b. The lift was in constant motion downwards. c. The lift while in downward motion suddenly stopped. d. The lift while in upward motion suddenly stopped.

A monkey of mass \(40 \mathrm{~kg}\) climbs on a massless rope of breaking strength \(600 \mathrm{~N}\). The rope will break if the monkey a. Climbs up with a uniform speed of \(5 \mathrm{~m} / \mathrm{s}\). b. Climbs up with an acceleration of \(6 \mathrm{~m} / \mathrm{s}^{2}\). c. Climbs down with an acceleration of \(4 \mathrm{~m} / \mathrm{s}^{2}\). d. Climbs down with a uniform speed of \(5 \mathrm{~m} / \mathrm{s}\).

When a body is stationary a. there is no force acting on it b. the forces acting on its are not in contact with it c. the combination of forces acting on it balance each other d. the body is in vacuum

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