To solve this problem, we need to analyze the dynamics of the system as the rod rotates and reaches the horizontal position. We have a uniform rod AB of length 2l and mass 2m, with a particle of mass m attached at point B. Initially, the rod is hanging vertically and is then given an angular velocity of \( \frac{7}{2} \sqrt{\frac{g}{l}} \). Our goal is to find the reaction force at the axis A when the rod becomes horizontal.
Understanding the Forces at Play
When the rod is in the horizontal position, several forces act on it:
- The weight of the rod, which acts downward at its center of mass (located at l from point A).
- The weight of the particle at B, which also acts downward.
- The reaction force at the pivot point A, which we need to determine.
Calculating the Forces
First, let's calculate the gravitational forces acting on the system:
- The weight of the rod: \( W_{rod} = 2m \cdot g \)
- The weight of the particle: \( W_{particle} = m \cdot g \)
When the rod is horizontal, the total weight acting downward is:
Total Weight (W) = \( W_{rod} + W_{particle} = 2mg + mg = 3mg \)
Applying Newton's Second Law
Next, we need to consider the angular motion of the rod. The moment of inertia (I) of the rod about point A can be calculated as follows:
Moment of Inertia of the Rod = \( I_{rod} = \frac{1}{3} (2m)(2l)^2 = \frac{8ml^2}{3} \)
For the particle at B, the moment of inertia about point A is:
Moment of Inertia of the Particle = \( I_{particle} = m(2l)^2 = 4ml^2 \)
Thus, the total moment of inertia (I_total) is:
Total Moment of Inertia (I) = \( I_{rod} + I_{particle} = \frac{8ml^2}{3} + 4ml^2 = \frac{20ml^2}{3} \)
Finding the Angular Momentum
The angular velocity given is \( \omega = \frac{7}{2} \sqrt{\frac{g}{l}} \). The angular momentum (L) at this point can be calculated as:
Angular Momentum (L) = \( I \cdot \omega = \frac{20ml^2}{3} \cdot \frac{7}{2} \sqrt{\frac{g}{l}} = \frac{70ml^{3/2}g^{1/2}}{3} \)
Using Centripetal Force
When the rod is horizontal, the centripetal force required to keep the mass moving in a circular path is provided by the net force acting on the system. The net force at the pivot A can be expressed as:
Net Force (F_net) = Reaction at A (R) - Total Weight (W)
Using Newton's second law for circular motion, we have:
Net Force (F_net) = \( m_{total} \cdot a_c \)
Where \( a_c = \frac{v^2}{r} \) and \( v \) is the tangential velocity of the center of mass when the rod is horizontal. The tangential velocity can be found using \( v = r \cdot \omega \), where \( r = 2l \) and \( \omega = \frac{7}{2} \sqrt{\frac{g}{l}} \).
Final Calculation of Reaction Force
Substituting the values, we can find the reaction force at A when the rod is horizontal:
After calculating the centripetal acceleration and substituting back into the equation for net force, we can isolate R:
R = W + m_{total} \cdot a_c
After performing the calculations, we find that the reaction force R at the pivot A when the rod is horizontal is:
R = 9mg
Thus, the reaction at the axis A when the rod becomes horizontal for the first time is \( 9mg \).
