Question

Course Contents> Set 03 (09/25 Tu 10 PM) Air puck on a table An air puck of mass 0.604 kg is tied to a string and allowed to revolve in a circle of radius 1.13 m on a frictionless horizontal table. in the fiqure. Timer□ Notes è Evaluate Feedback-Print The suspended mass remains in equilibrium while the puck on the tabletop revolves. What is the tension in the string? Submit Answer Tries 0/12 What is the magnitude of the force which causes the centripetal acceleration of the puck? Submit Answer Tries O/12 ehepg What is the speed of the puck? Subeit Answer Tries 0/12 Threaded View C Other ïewsm My general preferences on what is marked as NEW hronological View

Answer 1

M = 1.14 kg m = 0.604 kg r = 1.13 m g = 9.81 m/s^2 T = Mg = (1.14) (9.81) Therefore T = 11.18 N Tension in the string The centripetal force is equal to the tension in the string in magnitude Therefore F_CP = T = 11.18 N magnitude of force causing (centripetal acceleration) Therefore mv^2/r = T = > v^2 = Tr/m Therefore v = squareroot Mgr/m = squareroot (1.14) (9.81) (1.13)/(0.604) Therefore v = 4.574 m/s speed of the puck mu = 0.290 M_1 = ? M_2 = 3.50 kg theta = 15 degree N = M_1 g cos theta normal reaction f = mu N = mu M_1 g cos theta frictional force T = M_2 a Tension in the string M_1 a = T - M_1 g sin theta - f Therefore M_1a = M_2 a - M_1 g sin theta - mu M_1 g cos theta