Question

3. As shown on the right, a thick ring with a mass of 200.0 grams has very low mass spokes and a low mass central hub supporting it on a very low friction bearing. String of negligible mass is tied and wrapped around the outer edge so it can freely M, -200.0 grams R, 5.00 cm R, 10.00 em unwind, but not slip. Attached to the end of the string is a mass of 100.0 grams. The outer radius of the disk is 10.00 cm and the inner radius of 5.00 cm. If the hanging mass is released from rest, find M,-1000 grams its downward acceleration. (10 points) a=? Given: ck 1/2M(R2+R,) g 9.80 m/s

Answer 1

Net torque on thick ring mg + T = i = torque due to pull = torque due to tension = 1/2 m (R^2_1 + R^2_2) alpha^rightarrow = angular acceleration relation b/w angular acceleration and translation acceleration a^rightarrow = R_2 alpha^rightarrow Therefore 0 + (T^rightarrow times R^rightarrow_2) = i (a/R_2) rightarrow T R_2 sin 90 degree = 1/2 m_2 (R^2_1 + R^2_2) a/R_2 rightarrow T = 1/2 m_1 (R^2_1/R^2_2 + 1) a --(1) \r\n for mass m_2 F_net = F^rightarrow_m_2 g - F^rightarrow_T (from newton 2^nd law) rightarrow m_2 a = m_2 - T rightarrow m_2 a = m_2 g - 1/2 m ((R_1/R_2)^2 + s) a rightarrow a [m_2 + 1/2 m_1 (R^2_1/R_2)^2 + 1] = m_2 g rightarrow a [1 times 10^- 3 + 1/2 2 times 10^- 3 ((5/10)^2 + 1)] = 10^- 3 times 9.8 rightarrow a times 10^- 3 [1 + (0.25 + 1)] = 10^- 3 times 9.8 rightarrow [1 + 1.25] = 9.8