Two particles of masses m_1 and m_2 are connected by a light rigid rod of length r . The system rotates at an angular speed ω about an axis passing through the centre of mass of the system and perpendicular to the rod.

Question

Quantrex Academy · Accepted Answer

Correct answer: B. μ r^2 ω

Let the distances of the masses m_1 and m_2 from the centre of mass be r_1 and r_2 respectively. r_1 = (m_2 r)/(m_1 + m_2) and r_2 = (m_1 r)/(m_1 + m_2) The moment of inertia of the system about the axis passing through the centre of mass and perpendicular to the rod is I = m_1 r_1^2 + m_2 r_2^2 Substituting the values of r_1 and r_2 : I = m_1 ( (m_2 r)/(m_1 + m_2) )^2 + m_2 ( (m_1 r)/(m_1 + m_2) )^2 I = (m_1 m_2^2 r^2)/((m_1 + m_2)^2) + (m_2 m_1^2 r^2)/((m_1 + m_2)^2) I = (m_1 m_2 (m_1 + m_2) r^2)/((m_1 + m_2)^2) I = (m_1 m_2)/(m_1 + m_2) r^2 = μ r^2 The angular momentum of the system is given by L = I ω L = μ r^2 ω

Two particles of masses m_1 and m_2 are connected by a light rigid...

Options

Correct answer

Step-by-step solution

More from Rotational Mechanics