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Chapter 9: Problem 13

A coin is placed at the edge of a horizontal disc rotat about a vertical axis through its axis with a uniform ang?speed \(2 \mathrm{rad} \mathrm{s}^{-1} .\) The radius of the disc is \(50 \mathrm{~cm}\). Find the minimum coefficient of friction between disc and coin so that the coin does not slip \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\). (1) \(0.1\) (2) \(0.2\) (3) \(0.3\) (4) \(0.4\)

Short Answer

Expert verified
The minimum coefficient of friction is 0.2, corresponding to option (2).

Step by step solution

01

Understanding the Problem

A coin is placed on the edge of a rotating disc, and we need to find the minimum coefficient of friction that prevents the coin from slipping. The angular speed of the disc is given as \( \omega = 2 \text{ rad/s} \) and the radius \( r = 50 \text{ cm} = 0.5 \text{ m} \). We're also given \( g = 10 \text{ m/s}^2 \).
02

Calculate Centripetal Force

The centripetal force \( F_c \) required to keep the coin moving in a circle is given by \( F_c = m \omega^2 r \) where \( m \) is the mass of the coin. We don't need the mass for the friction coefficient but let's write the equation. Substitute \( \omega = 2 \text{ rad/s} \) and \( r = 0.5 \text{ m} \) to get \( F_c = m (2)^2 (0.5) = 2m \).
03

Determine Frictional Force Requirement

The frictional force \( F_f \) that prevents the coin from slipping is \( \mu mg \), where \( \mu \) is the coefficient of friction. For the coin to not slip, the frictional force must be equal to the centripetal force: \( \mu mg = 2m \).
04

Solve for Minimum Coefficient of Friction

By equating the frictional and centripetal forces \( \mu mg = 2m \), we can solve for \( \mu \): \( \mu = \frac{2}{g} = \frac{2}{10} = 0.2 \).
05

Verify the Answer

The given problem options include 0.1, 0.2, 0.3, and 0.4. Our calculated minimum \( \mu \) is 0.2, which matches option (2).

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Key Concepts

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

Centripetal Force
Centripetal force plays a vital role in rotational motion, ensuring that objects traveling in a circular path remain on that path. When something moves in a circle, it is constantly changing direction, meaning it is accelerating even if the speed is constant. This requires a force directed towards the center of the circle, known as centripetal force.
  • The centripetal force keeps the coin from flying off the edge of the disc.
  • It pulls the object towards the center, ensuring circular motion is maintained.
In the exercise, the formula for centripetal force is given as \( F_c = m\omega^2 r \). Here, \( m \) is the mass, \( \omega \) is the angular speed, and \( r \) is the radius of the circle.
For our coin, even though the mass cancels out in the final calculation for the coefficient of friction, this force is crucial for ensuring the coin remains on the disc.
Coefficient of Friction
The coefficient of friction is a measure of the force resisting the relative motion between two surfaces in contact. It determines how 鈥榞rippy鈥� a surface is.
  • A higher coefficient means more grip, preventing sliding.
  • Frictional force can be calculated as \( F_f = \mu mg \), where \( \mu \) is the coefficient of friction.
For our problem with the disc and coin, the friction between the disc and the coin is necessary to counteract the outward force from the coin's circular motion. If the frictional force is not enough, the coin would slip off.
By balancing the centripetal force and the frictional force, you can determine the minimum coefficient needed. We derived \( \mu = \frac{2}{g} = 0.2 \), aligning perfectly with the provided answer options.
Angular Speed
Angular speed describes how fast an object rotates or revolves relative to another point. It's measured in radians per second (rad/s), and in this exercise, the disc has a uniform angular speed of \(2 \text{ rad/s}\).
  • Angular speed determines how quickly the disc spins.
  • It's an essential factor in calculating centripetal force, as seen in \( F_c = m\omega^2 r \).
Using the given angular speed, we could calculate the necessary centripetal force, taking into account the disc's radius. This understanding helps us find how much friction is needed to keep the coin in place. Understanding angular speed and its contribution to rotational motion is fundamental when solving such physics problems.

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Most popular questions from this chapter

A particle starts from rest and performing circular motion of constant radius with speed given by \(v=\alpha \sqrt{x}\) where \(\alpha\) is a constant and \(x\) is the distance covered. The correct graph of magnitude of its tangential acceleration \((a)\) and centripetal acceleration \(\left(a_{e}\right)\) versus \(t\) will be (1) (2) CC1CCC(C)(C)C1 (3) (4)

A car moving along a circular track of radius \(50.0 \mathrm{~m}\) accelerates from rest at \(3.00 \mathrm{~ms}^{-2}\). Consider a situation when the car's centripetal acceleration equals its tangential acceleration. Then (1) The angle around the track does the car travel is 1 rad (2) The magnitude of the car's total acceleration at that instant is \(3 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (3) Time elapses before this situation is \(\sqrt{\frac{50}{3}} \mathrm{~s}\). (4) The distance travelled by the car during this time is \(25 \mathrm{~m}\)

A particle moves in a circle of radius \(20 \mathrm{~cm}\). Its linear speed is given by \(v=2 t\) where \(t\) is in seconds and \(v\) in \(\mathrm{ms}^{-1}\) Then (1) The radial acceleration at \(t=2 \mathrm{~s}\) is \(80 \mathrm{~m} \mathrm{~s}^{-2}\). (2) Tangential acceleration at \(t=2 \mathrm{~s}\) is \(2 \mathrm{~ms}^{-2} .\) (3) Net acceleration at \(t=2 \mathrm{~s}\) is greater than \(80 \mathrm{~ms}^{-2}\). (4) Tangential acceleration remains constant in magnitude.

Two strings of length \(l=0.5 \mathrm{~m}\) each are connected to a block of mass \(m=2 \mathrm{~kg}\) at one end and their ends are attached to the point \(A\) and \(B, 0.5 \mathrm{~m}\) apart on a vertical pole which rotates with a constant angular velocity \(\omega=7 \mathrm{rad} / \mathrm{s}\). Find the ratio \(T_{1} / T_{2}\) of tension in the upper string \(\left(T_{1}\right)\) and the lower string \(\left(T_{2}\right)\). [Use \(\left.g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right]\)

Two particles describe the same circle of radius \(R\) in the same direction with the same speed \(v\), then at the given instant relative angular velocity of 2 with respect to 1 will be (1) \(\frac{2 v \sin \frac{\theta}{2}}{R}\) (2) \(\frac{v}{2 R \sin \frac{\theta}{2}}\) (3) \(\frac{v}{R}\) (4) \(\frac{v \cos \frac{\theta}{2}}{R}\)
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