Question

9. (10 points) Two masses, m1 and m2 are connected by a massless chord over a disc of radius R and mass M as shown in the figure. m1 m2. The chord does not slip on the disk, which can turn on a frictionless bearing. The masses are then released. a) What is the angular acceleration of the disc? b) What is the ratio of the kinetic energy of the disc to the total kinetic energy of the two masses m1 and m2.

Answer 1

At Block m_1 m_1 g - T_1 = m_1a rightarrow 1 At Block m_2 T_2 - m_2 g = m_2 a rightarrow 2 At Disc (T_1 - T_2) R = I alpha (T_1 - T_2) R = 1/2 MR^2 times a/R T_1 - T_2 = 1/2 ma rightarrow 3 from 1 , 2 and 3 m_1 g - m_2 g = m_1a + m_2 a + 1/2 Ma therefore a = (m_1 - m_2) g/m_1 + m_2 + M/2 kinetic energy of Disc = 1/2 Iw^2 = 1/2 (1/2 MR^2) w^2 = 1/4 MV^2 kinetic energy of masses = 1/2 m_1 v^2 + 1/2 m_2 v^2 = 1/2 (m_1 + m_2) v^2 (KE)_Disc/(KE)_masses = 1/4MV^2/1/2 (m_1 + m_2)v^2 = M/2 (m_1 + m_2)