A block of mass \(10 kg\) is in contact against the inner wall of a hollow cylindrical drum of radius \( 1 m.\) The coefficient of friction between the block and the inner wall of the cylinder is \( 0.1\). The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be: (\(g=10 \frac{m}{s^2}\)

Question

Let ω1, ω2 and ω3 be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If x1, x2 and x3 are their respective angular distances in 1 minute then the factor which remains constant (k) is

Answer 1

To solve this problem, we need to determine the minimum angular velocity required to keep a block stationary against the inner wall of a rotating cylindrical drum. The block of mass \(10 \, \text{kg}\) is subjected to centrifugal force, friction, and gravitational force. Let's calculate the minimum angular velocity step by step.

First, identify the forces acting on the block:
- Centrifugal Force: This force is provided by the rotation of the cylinder and is directed outward. It can be expressed as \( F_c = m \cdot r \cdot \omega^2 \), where \( m = 10 \, \text{kg} \) is the mass of the block, \( r = 1 \, \text{m} \) is the radius of the drum, and \( \omega \) is the angular velocity.
- Gravitational Force: This force acts downward and is given by \( F_g = m \cdot g = 10 \cdot 10 = 100 \, \text{N} \).
- Frictional Force: This is the force that opposes the motion, enabling the block to stay in place. It is given by \( F_f = \mu \cdot N \), where \( \mu = 0.1 \) is the coefficient of friction, and \( N \) is the normal force, which equals the centrifugal force \( F_c \) in this setup.
For the block to remain stationary, the frictional force should be at least equal to the gravitational force. Therefore: \( F_f = F_g \). Substituting the friction formula, we get: \[ \mu \cdot N = m \cdot g \] Since \( N = F_c = m \cdot r \cdot \omega^2 \), we can substitute: \[ \mu \cdot (m \cdot r \cdot \omega^2) = m \cdot g \]
Solving for the angular velocity \( \omega \): \[ \mu \cdot m \cdot r \cdot \omega^2 = m \cdot g \] \[ \omega^2 = \frac{g}{\mu \cdot r} \] \[ \omega^2 = \frac{10}{0.1 \cdot 1} = \frac{10}{0.1} = 100 \] \[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]
Thus, the minimum angular velocity required to keep the block stationary is \( \omega = 10 \, \text{rad/s} \).

The correct answer is the option: \(10 \, \text{rad/s}\).

Answer 2

To solve this problem, we need to determine the minimum angular velocity required to keep a block stationary against the inner wall of a rotating cylindrical drum. The block of mass \(10 \, \text{kg}\) is subjected to centrifugal force, friction, and gravitational force. Let's calculate the minimum angular velocity step by step.

First, identify the forces acting on the block:
- Centrifugal Force: This force is provided by the rotation of the cylinder and is directed outward. It can be expressed as \( F_c = m \cdot r \cdot \omega^2 \), where \( m = 10 \, \text{kg} \) is the mass of the block, \( r = 1 \, \text{m} \) is the radius of the drum, and \( \omega \) is the angular velocity.
- Gravitational Force: This force acts downward and is given by \( F_g = m \cdot g = 10 \cdot 10 = 100 \, \text{N} \).
- Frictional Force: This is the force that opposes the motion, enabling the block to stay in place. It is given by \( F_f = \mu \cdot N \), where \( \mu = 0.1 \) is the coefficient of friction, and \( N \) is the normal force, which equals the centrifugal force \( F_c \) in this setup.
For the block to remain stationary, the frictional force should be at least equal to the gravitational force. Therefore: \( F_f = F_g \). Substituting the friction formula, we get: \[ \mu \cdot N = m \cdot g \] Since \( N = F_c = m \cdot r \cdot \omega^2 \), we can substitute: \[ \mu \cdot (m \cdot r \cdot \omega^2) = m \cdot g \]
Solving for the angular velocity \( \omega \): \[ \mu \cdot m \cdot r \cdot \omega^2 = m \cdot g \] \[ \omega^2 = \frac{g}{\mu \cdot r} \] \[ \omega^2 = \frac{10}{0.1 \cdot 1} = \frac{10}{0.1} = 100 \] \[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]
Thus, the minimum angular velocity required to keep the block stationary is \( \omega = 10 \, \text{rad/s} \).

The correct answer is the option: \(10 \, \text{rad/s}\).

The Correct Option is C

Solution and Explanation

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