A mass m1 is connected by a light string that passes over a pulley of mass M to a mass m2 sliding on a frictionless horizontal surface as shown in the figure. There is no slippage between the string and the pulley. The pulley has a radius of 25.0 cm and a moment of inertia of ½ MR2. If m1 is 4.00 kg, m2 is 4.00 kg, and M is 4.00 kg, then what is the tension in the string attached to m2?
Question
A mass m1 is connected by a light string that passes over a pulley of mass M to a mass m2 sliding on a frictionless horizontal surface as shown in the figure. There is no slippage between the string and the pulley. The pulley has a radius of 25.0 cm and a moment of inertia of ½ MR2. If m1 is 4.00 kg, m2 is 4.00 kg, and M is 4.00 kg, then what is the tension in the string attached to m2?
Solution
To solve this problem, we need to use the principles of physics, specifically Newton's second law and the concept of torque.
Step 1: Identify the forces acting on m2. The only forces acting on m2 are the tension in the string (T) and the force of gravity (m2g). Since m2 is not accelerating vertically, these forces must balance out, so T = m2g.
Step 2: Identify the forces acting on m1. The forces acting on m1 are the tension in the string (T) and the force of gravity (m1g). Since m1 is accelerating downwards, the net force on m1 is m1g - T.
Step 3: Identify the forces acting on the pulley. The tension in the string exerts a torque on the pulley, causing it to rotate. The torque τ is given by τ = I*α, where I is the moment of inertia of the pulley and α is its angular acceleration. Since the string does not slip on the pulley, the angular acceleration of the pulley is related to the linear acceleration of m1 by α = a/R, where R is the radius of the pulley.
Step 4: Write down Newton's second law for m1, m2, and the pulley. For m1, this gives m1g - T = m1a. For m2, this gives T = m2g. For the pulley, this gives τ = Iα = I*a/R.
Step 5: Solve these equations simultaneously to find the tension T. Substituting the expression for T from the equation for m2 into the equation for m1 gives m1g - m2g = m1a. Solving for a gives a = (m1g - m2g)/m1. Substituting this into the equation for the pulley gives Ia/R = TR, or T = Ia/R^2.
Step 6: Substitute the given values into this equation to find T. The moment of inertia I is given by I = 1/2MR^2, so T = (1/2MR^2)((m1g - m2g)/m1)/R^2 = (1/2M*(m1g - m2g))/m1.
Given that m1 = m2 = M = 4.00 kg, g = 9.81 m/s^2, and R = 0.25 m, this gives T = (1/24.00 kg(4.00 kg9.81 m/s^2 - 4.00 kg9.81 m/s^2))/4.00 kg = 0 N.
So, the tension in the string attached to m2 is 0 N.
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