An Atwoods machine is made with masses M1=100g, M2= 120g, and pulley mass of 100

Question

An Atwoods machine is made with masses M1=100g, M2= 120g, and pulley mass of 100g. What is the percent error in acceleration (relative to the corrected formula) calculated by neglecting the pulley mass versus incorporating the pulley mass?

Explanation / Answer

without taking the mass of the pulley into account, your expression for acceleration is a = (m2-m1)g/(m2+m1) taking the mass of the pulley into account, our three equations become: for mass m2, m2>m1: T2 - m2g = - m2a (the negative sign means m2 is accelerating down) for mass m1: T1 - m1g = m1a for the pulley torque = (T2-T1)R = I alpha I = 1/2 MR^2 where M, R are the mass and radius of the pulley, and 1/2 MR^2 is the moment of inertia for a solid disk, alpha is the angular accel = a/R therefore the torque equation becomes (T2-T1)R = 1/2 MR^2(a/R) T2-T1= 1/2 Ma combine these equations to get your expression for acceleration: a=(m2-m1)g/(m1+m2+1/2M) the two accelerations are easily calculated: a(neglecting pulley) = 0.89m/s/s a(with pulley) = 0.61m/s/s percent error = 31.3%

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