Concepts Of Physics MCQ Edition [Volume 1]PhysicsSimple Harmonic Motion
Two small balls, each having a mass m , are joined by a light rigid rod of length L . This system is suspended from its centre using a thin wire with a torsional constant k . The rod is turned about the wire by an angle heta₀ and released. Determine the force exerted by the rod on one of the balls as the system crosses the mean position.
Options
- A[ k^2 ₀^2 L^2 + m^2 g^2 ]^ 1/2
- B[ 4k^2 ₀^4 L^2 + m^2 g^2 ]^ 1/2
- C[ k^2 ₀^4 4L^2 + m^2 g^2 ]^ 1/2
- D[ k^2 ₀^4 L^2 + m^2 g^2 ]^ 1/2
Correct answer
D. [ k^2 ₀^4 L^2 + m^2 g^2 ]^ 1/2
Step-by-step solution
The moment of inertia of the system about the suspension wire is given by the sum of the moments of inertia of the two balls. I = m ( L 2 )^2 + m ( L 2 )^2 = mL^2 2 By the principle of conservation of mechanical energy, the maximum potential energy stored in the torsional wire is converted into kinetic energy as the system crosses the mean position. 1 2 k ₀^2 = 1 2 I ^2 Substituting the value of I : k ₀^2 = ( mL^2 2 ) ^2 ^2 = 2k ₀^2 mL^2 As the system crosses the mean position, the ball experiences two forces exert