a 4.00kg counter weight is attatched to a light cord that swound around a reel.

Question

a 4.00kg counter weight is attatched to a light cord that swound around a reel. The reel is a uniform solid cylinder of radius8.00cm and a mass 2.00kg. (a) what is the net torque on the systemabout the point O? (b) When the counter weight has a speed of = v/R. Determine the total angular momentum of the systemabout O. (c) Using that = dL/dt and your result from part(b), calculate the acceleration of the counterweight. a 4.00kg counter weight is attatched to a light cord that swound around a reel. The reel is a uniform solid cylinder of radius8.00cm and a mass 2.00kg. (a) what is the net torque on the systemabout the point O? (b) When the counter weight has a speed of = v/R. Determine the total angular momentum of the systemabout O. (c) Using that = dL/dt and your result from part(b), calculate the acceleration of the counterweight.

Explanation / Answer

The mass of the counter weight attached to a light cord thatis wound around a reel is m = 4.00 kg The reel is a uniform solid cylinder of radius r = 8.00cm anda mass m1= 2.00kg. (a)The net torque on the system about the point O is = F * r Here,F = m * g or = m * g * r Here,g = 9.8 m/s2 or = 4.00 * 9.8 * 8.00 * 10-2 or = 3.136 N-m (b)The net torque on the system about the point O is = I * w or w = (/I) Here,I = (m * r2/2) or w = (/(m * r2/2)) = (2/m *r2) Substituting the values in the above equation,we get w = (2 * 3.136/4.00 * (8.00 *10-2)2) or w = 245 rad/s The total angular momentum of the system about O is L = m * v * r = m * r2 * w or L = 4.00 * (8.00 * 10-2)2 * 245 or L = 6.272 kg.m/s (c)Using that = dL/dt and the result from part (b),theacceleration of the counterweight is a = (v2/r) = ((r * w)2/r) = r *w2 or a = 8.00 * 10-2 * (245)2 or a = 4802 m/s2 or a = 4802 m/s2

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