Physics Analysis: Coin on Rotating Disc
The problem asks for the coefficient of friction ($\mu_s$) required to keep a coin stationary on a horizontally rotating disc. The coin experiences a centripetal force due to the rotation, which must be balanced by the static friction force between the coin and the disc.
Physics Principles and Formulas
For the coin to remain at rest, the static friction force ($f_s$) must provide the necessary centripetal force ($F_c$).
- Centripetal Force: $ F_c = m a_c $, where $m$ is the mass of the coin and $a_c$ is the centripetal acceleration.
- Centripetal Acceleration: $ a_c = \omega^2 r $, where $\omega$ is the angular velocity and $r$ is the radius.
- Angular Velocity: $ \omega = 2 \pi f $, where $f$ is the frequency of rotation.
- Maximum Static Friction: $ f_{s,max} = \mu_s N $, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.
- Normal Force on a horizontal surface: $ N = mg $, where $g$ is the acceleration due to gravity.
The condition for the coin to remain at rest is $F_c \le f_{s,max}$. For the minimum required coefficient of friction, we consider the case where $F_c = f_{s,max}$.
$ m a_c = \mu_s N $ $ m (\omega^2 r) = \mu_s (mg) $ $ \mu_s = \frac{\omega^2 r}{g} $Calculation Steps
- Calculate the angular velocity ($\omega$): Given $f = 3.5$ rev/s. $ \omega = 2 \pi f = 2 \pi (3.5 \text{ s}^{-1}) = 7 \pi \text{ rad/s} $.
- Substitute values into the formula for $\mu_s$: Given $r = 1.25$ cm $= 0.0125$ m and $g = 10$ m/s². $ \mu_s = \frac{(7 \pi \text{ rad/s})^2 \times (0.0125 \text{ m})}{10 \text{ m/s}^2} $ $ \mu_s = \frac{49 \pi^2 \times 0.0125}{10} $
- Approximate and calculate $\mu_s$: Using the approximation $ \pi^2 \approx 10 $: $ \mu_s \approx \frac{49 \times 10 \times 0.0125}{10} $ $ \mu_s \approx 49 \times 0.0125 $ $ \mu_s \approx 0.6125 $
- Determine the answer: The calculated coefficient of friction is approximately $0.6125$. This value is closest to the option $0.6$.
Similar Questions
- Two masses $m$ and $2m$ are connected by a light string going over a pulley (disc) of mass $30m$ with radius $r=0.1 \text{ m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The $2m$ mass is released from rest and its speed when it has descended through a height of 3.6 m is _________ m/s. (Assume string does not slip and $g = 10 \text{ m/s}^2$)
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($g = \text{gravitational acceleration}$)
- A soap bubble of surface tension $0.04 \text{ N/m}$ is blown to a diameter of $7 \text{ cm}$. If $(15000 - x) \text{ }\mu\text{J}$ of work is done in blowing it further to make its diameter $14 \text{ cm}$, then the value of $x$ is _______.
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Important Questions from Mechanics
- Two masses $m$ and $2m$ are connected by a light string going over a pulley (disc) of mass $30m$ with radius $r=0.1 \text{ m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The $2m$ mass is released from rest and its speed when it has descended through a height of 3.6 m is _________ m/s. (Assume string does not slip and $g = 10 \text{ m/s}^2$)
- In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $= 0.05 \text{ mm}$, then the least count of the vernier callipers is _______ mm.
- A flexible chain of mass $m$ hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is $30^\circ$. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is _______.
A thin uniform rod ($X$) of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is __________.
($g = \text{gravitational acceleration}$)- A soap bubble of surface tension $0.04 \text{ N/m}$ is blown to a diameter of $7 \text{ cm}$. If $(15000 - x) \text{ }\mu\text{J}$ of work is done in blowing it further to make its diameter $14 \text{ cm}$, then the value of $x$ is _______.
($\pi = 22/7$)
