a cylinder of mass m is kept on the edge of a plank of mass 2m and length 12m which inturn is kept on smooth ground.coefficient of friction btw the cylinder n plank is 0.1. the cylinder is given an impulse which imparts it a velocity 7m/s bt no angular velocity.find the time after which the cylinder falls off the plank

To determine the time after which the cylinder falls off the plank, we need to analyze the motion of both the cylinder and the plank. The key here is to understand how the impulse given to the cylinder affects its motion relative to the plank, especially considering the friction between them.

Understanding the System

We have a cylinder of mass \( m \) on a plank of mass \( 2m \). The plank is resting on a smooth surface, which means there’s no friction between the plank and the ground. The coefficient of friction between the cylinder and the plank is \( 0.1 \). When the cylinder receives an impulse, it gains a velocity of \( 7 \, \text{m/s} \) without any angular velocity.

Analyzing Forces and Motion

When the cylinder is given an impulse, it starts moving to the right with a velocity of \( 7 \, \text{m/s} \). The friction between the cylinder and the plank will act to the left on the cylinder, trying to slow it down relative to the plank. The frictional force can be calculated using the formula:

• Frictional Force (F_f) = \( \mu \cdot N \)

Here, \( \mu \) is the coefficient of friction (0.1), and \( N \) is the normal force, which equals the weight of the cylinder, \( mg \). Therefore:

• F_f = \( 0.1 \cdot mg \)

Calculating the Acceleration of the Cylinder

The frictional force will cause the cylinder to decelerate. Using Newton's second law, we can find the acceleration \( a \) of the cylinder:

• Acceleration (a) = \( \frac{F_f}{m} = \frac{0.1 \cdot mg}{m} = 0.1g \)

Substituting \( g \approx 9.81 \, \text{m/s}^2 \), we find:

• a = \( 0.1 \cdot 9.81 \approx 0.981 \, \text{m/s}^2 \)

Finding the Relative Motion

The plank will also start moving due to the frictional force acting on it. The frictional force acting on the plank is equal to the frictional force acting on the cylinder but in the opposite direction. Thus, the acceleration of the plank \( a_p \) can be calculated as:

• Acceleration of the Plank (a_p) = \( \frac{F_f}{2m} = \frac{0.1 \cdot mg}{2m} = 0.05g \)

Calculating this gives:

• a_p = \( 0.05 \cdot 9.81 \approx 0.4905 \, \text{m/s}^2 \)

Relative Velocity and Time Calculation

Now, we need to find the time \( t \) when the cylinder falls off the plank. The relative velocity \( v_r \) between the cylinder and the plank will change over time due to their respective accelerations:

• Initial Relative Velocity (v_{r0}) = \( 7 \, \text{m/s} \)
• Relative Acceleration (a_r) = \( a - a_p = 0.981 - 0.4905 = 0.4905 \, \text{m/s}^2 \)

The cylinder will fall off the plank when the distance it travels relative to the plank equals the length of the plank, which is \( 12 \, \text{m} \). The distance covered under uniform acceleration can be calculated using the equation:

• d = \( v_{r0} t + \frac{1}{2} a_r t^2 \)

Setting \( d = 12 \, \text{m} \), we have:

• \( 12 = 7t + \frac{1}{2} (0.4905) t^2 \)

This is a quadratic equation in the form of:

• \( 0.24525 t^2 + 7t - 12 = 0 \)

Solving the Quadratic Equation

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 0.24525 \), \( b = 7 \), and \( c = -12 \):

• \( b^2 - 4ac = 7^2 - 4 \cdot 0.24525 \cdot (-12) \)
• \( = 49 + 11.706 = 60.706 \)
• \( t = \frac{-7 \pm \sqrt{60.706}}{2 \cdot 0.24525} \)

Calculating this gives two potential solutions for \( t \). We only consider the positive root since time cannot be negative. After evaluating, we find:

• \( t \approx 2.5 \, \text{s} \)

Thus, the cylinder will fall off the plank after approximately \( 2.5 \) seconds. This analysis illustrates how the interplay of forces and motion leads to the outcome of the cylinder's trajectory relative to the plank.