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Introductory Physics Homework Help
Second Newton's law in rotation with pulley.
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[QUOTE="PhysicsKush, post: 5479868, member: 582643"] [h2]Homework Statement [/h2] The rope doesn't slide on the pulley, it's mass is negligible and it doesn't stretch .The pulley's moment of inertia is 0.00300 kg * m^2 with a radius of 10 cm. As the first mass goes down the rope's friction generates a moment of force of 0.150 N*m opposed to the angular speed of the pulley. Initially , both masses are motionless and m1 is at a height of 1.5m(h = 1.5m). m[SUB]1[/SUB] = 1.5 kg m[SUB]2[/SUB] = 0.8 kg [URL]https://gyazo.com/074104f2af4ac27811136c21b4e1bcfb[/URL][h2]Homework Equations[/h2] τ = r*F τ[SUB]resultant[/SUB] = Σ(τ) = I* α[SUB]z[/SUB] F[SUB]resultant[/SUB] = ma α[SUB]z[/SUB] = a[SUB]θ[/SUB] / r -a[SUB]y1[/SUB] = a[SUB]y2[/SUB][h2]The Attempt at a Solution[/h2] T[SUB]1[/SUB] = -m[SUB]1[/SUB]a[SUB]y1[/SUB] + m[SUB]1[/SUB]g T[SUB]2[/SUB] = m[SUB]2[/SUB]a[SUB]y2[/SUB] + m[SUB]2[/SUB]g -T[SUB]1[/SUB] + T[SUB]2[/SUB] - 1.5 = I*a[SUB]y1[/SUB] replacing the values in the last equation with the first two equations and isolating a results 3.69 m/s[SUP]2[/SUP] the real solution is a[SUB]1[/SUB] = -2.06 m/s[SUP]2[/SUP] [/QUOTE]
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Introductory Physics Homework Help
Second Newton's law in rotation with pulley.
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