Two metal disks, one with radius = 2.57 and mass = 0.900 and the other with radi

Question

Two metal disks, one with radius = 2.57 and mass = 0.900 and the other with radius = 4.97 and mass = 1.63 , are welded together and mounted on a frictionless axis through their common center. A light string is wrapped around the edge of the smaller disk and a 1.51block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released? Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk.

Explanation / Answer

Two metal discs, one with radius R1 = 2.50 cm and mass M1 = 0.80 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 kg, are welded together and mounted on a frictionless axis through their common centre. a) What is the total moment of inertia of the two discs? b) A light string is wrapped around the edge of the smaller disc, and a 1.50-kg block is suspended from the free end of the string. If the block is released from rest a distance of 2.00 m from the floor, what is the speed just before it strikes the floor? Am I correct to assume that the total moment of inertia is the sum of the moment inertia of both discs? If the discs have a common axis, then the moments can be directly summed. The moment of a single disc is m*r²/2, so the moment of the two discs is 0.8*0.025²/2 + 1.60*0.05²/2; I = 0.00225 kg-m² Let Ft be the tension in the spring. The force balance on the hanging block (mass mb) is mb*g - Ft = mb*a The angular acceleration of the discs is a = T/I where T = torque on the discs. T = Ft*r so a = Ft*r/I but a = a*r a = Ft*r²/I. Solve the above eq for Ft and insert; Ft = mb*g = mb*a a = (mb*g - mb*a)*r²/I = g*mb*r²/I - mb*a*r²/I a*(1 + mb*r²/I) = mb*r²/I a = mb*r²/I / (1 + mb*r²/I) = 2.88 m/s² h = 0.5*a*t² so the time to reach the floor is v[2*h/a]. The speed is a*t, so v - v[2*h*a] = 3.4 m/s Alternate (and simpler) approach: Initial energy of the system is mb*g*h Final energy is 0.5*mb*v² + 0.5*I*?² ? = v/r so mb*g*h = 0.5*mb*v² + 0.5*I*v²/r² = v²*( 0.5*mb + 0.5*I/r²) v = v[2*mb*g*h/(mb + I/r²)] v = 3.4 m/s

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