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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Quizplus · Accepted Answer

First, we need to find the acceleration of the system. Starting with the two masses, we can use Newton's second law: m₁g − T = m₁a T − m₂g = m₂a where T is the tension in the string and a is the acceleration of the system. Next, we need to use the fact that there is no slippage between the string and the pulley, which means the linear velocity of the string is the same as the linear velocity of the pulley at the point of contact. We can relate this velocity to the angular velocity of the pulley using the radius of the pulley: v = ωR where v is the linear velocity, ω is the angular velocity, and R is the radius of the pulley. Taking the time derivative of this equation gives us: a = Rα where α is the angular acceleration of the pulley. We can relate the angular acceleration of the pulley to the tension in the string using the moment of inertia of the pulley: τ = Iα where τ is the torque on the pulley and I is the moment of inertia of the pulley. The torque on the pulley comes from the tension in the string, so we can write: τ = TR where T is the tension in the string. Putting it all together, we have: m₁g − T = m₁(Rα) T − m₂g = m₂(Rα) TR = Iα Using algebraic manipulation, we can solve for the tension in the string: T = (m₁ − m₂)g/(m₁ + m₂ + (I/R^2)) Plugging in the given values, we get: T = (4.00 kg − 3.00 kg) × 9.81 m/s^2/(4.00 kg + 3.00 kg + (MR²/0.30 m)^2 × 6.00 kg) T = 31.7 N

A Mass M1 Is Connected by a Light String That

A Mass M₁ Is Connected by a Light String That