Question no 9..

A cabin is accelerating up with gm/s.a pendulum of lengthy √3 is hanging from wall .find minimum velocity for completing the circle

To solve the problem of a pendulum hanging in a cabin that is accelerating upwards, we need to analyze the forces acting on the pendulum and the conditions required for it to complete a circular motion. The key here is to determine the minimum velocity at the top of the circular path that the pendulum must have to maintain tension in the string and not fall away from the circular path.

Understanding the Forces at Play

When the cabin accelerates upwards with an acceleration of \( g \) (where \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \)), the effective gravitational force acting on the pendulum changes. The pendulum experiences two forces: the gravitational force acting downwards and the tension in the string acting upwards.

Effective Gravity

In the accelerating frame of the cabin, the effective gravitational acceleration \( g_{\text{eff}} \) can be expressed as:

• \( g_{\text{eff}} = g + a \)

Since the cabin is accelerating upwards with \( a = g \), we have:

• \( g_{\text{eff}} = g + g = 2g \)

Condition for Completing the Circle

For the pendulum to complete a circular motion, at the highest point of the swing, the tension in the string must be non-negative. The centripetal force required to keep the pendulum moving in a circle is provided by the gravitational force acting on it. At the top of the circle, the forces can be summarized as:

• \( T + mg = \frac{mv^2}{L} \)

Where:

• \( T \) is the tension in the string (which can be zero at the minimum condition),
• \( m \) is the mass of the pendulum bob,
• \( v \) is the velocity at the top of the circle, and
• \( L \) is the length of the pendulum (in this case, \( L = \sqrt{3} \)).

Setting Up the Equation

At the minimum velocity condition, we can set the tension \( T \) to zero:

• \( mg = \frac{mv^2}{L} \)

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

• \( g = \frac{v^2}{L} \)

Now, substituting \( L = \sqrt{3} \):

• \( g = \frac{v^2}{\sqrt{3}} \)

Rearranging gives us:

• \( v^2 = g \cdot \sqrt{3} \)

Taking the square root of both sides, we find:

• \( v = \sqrt{g \cdot \sqrt{3}} \)

Calculating the Minimum Velocity

Now, substituting \( g \approx 9.81 \, \text{m/s}^2 \):

• \( v = \sqrt{9.81 \cdot \sqrt{3}} \)

Calculating this gives:

• \( v \approx \sqrt{9.81 \cdot 1.732} \approx \sqrt{17.00} \approx 4.12 \, \text{m/s} \)

Final Result

Therefore, the minimum velocity required for the pendulum to complete the circular motion while the cabin accelerates upwards at \( g \) is approximately \( 4.12 \, \text{m/s} \).