An object of mass m1 = 4.60 kg is connected by a light cord to an object of mass

Question

An object of mass m1 = 4.60 kg is connected by a light cord to an object of mass m2 = 3.00 kg on a frictionless surface (see figure). The pulley rotates about a frictionless axle and has a moment of inertia of 0.480 kg

Explanation / Answer

et the acceleration of the whole system = a T1 = (m1g - m1a) T2 = m2a the torque on the pulley is F*R where F = T1 - T2 --> torque = T1*R - T2*R ---> torque = (m1g - m1a)*R - m2a*R and in addition is torque = I*a = I*a/R with a = a/R since the tangential acceleration of the pulley is equal to the tangential acceleration of the whole system, we can write (m1g - m1a)*R - m2a*R = I*a/R (m1g - m1a)*R^2 - m2a*R^2 = I*a now solve for a m1gR^2 - m1aR^2 - m2aR^2 = I*a a(- m1R^2 - m2R^2 - I) = - m1gR^2 a = m1gR^2/(m1R^2 - m2R^2 - I) --> divide by R^2 a = m1g/(m1 + m2 + I/R^2) put in the given: a = 4.3*9.81/(4.3 + 3 + 0.42/0.28^2) a = 3.33274 m/s^2

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