A mass m₁ is connected by a light string that passes over a pulley of mass M to a mass m₂ as shown in the figure.Both masses move vertically and there is no slippage between the string and the pulley.The pulley has a radius of 30.0 cm and a moment of inertia of MR².If m₁ is 4.00 kg,m₂ is 3.00 kg and M is 6.00 kg,then what is the tension in the string that is attached to m₂? A) 20.7 N B) 25.5 N C) 31.7 N D) 35.2 N E) 41.3 N

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First, we need to draw free-body diagrams for both masses. For m₁, the tension in the rope pulls it upward, while gravity pulls it downward. For m₂, the tension in the rope pulls it upward, while gravity pulls it downward with a force of mg. Next, we need to set up equations of motion for each mass, using Newton's second law. For m₁: T - m₁g = m₁a For m₂: m₂g - T = m₂a We also need to use the fact that the acceleration of both masses is the same, and related to the angular acceleration of the pulley: a = R alpha where R is the radius of the pulley (0.30 m). Next, we need to consider the motion of the pulley itself. Since it is not a point mass, we need to use rotational dynamics: T R - I alpha = (1/2) M R^2 a where I is the moment of inertia of the pulley. Finally, we need to eliminate the acceleration (a) and angular acceleration (alpha) terms from these equations, to solve for T. We can do this by adding the first two equations: m₂g - m₁g = (m₂ + m₁) a and substituting for a in the equation for the pulley: T R - I (T/R) = (1/2) M R^2 [ (m₂ - m₁) / (m₂ + m₁) ] g Solving for T gives: T = [ I (m₂ - m₁) / (2 R^2) + (m₁ + m₂) g / 2 ] Plugging in the given values gives: T = (0.5 MR^2 + m₁ m₂ R^2 / (m₁ + m₂) ) g / (m₂ + m₁) T = (0.5 x 6.00 x 0.30^2 + 4.00 x 3.00 x 0.30^2 / 7.00) x 9.81 T = 31.7 N

A Mass M1 Is Connected by a Light String That

A Mass M₁ Is Connected by a Light String That