Hello, can you please show formulas and the way you worked out the problem? Why

Question

Hello, can you please show formulas and the way you worked out the problem? Why I am confused is because time was not given, nor revolutions per second or 60 seconds. No time. A uniform disk with mass 40kg and radius 0.2meters rotates about a frictionless axle. The disk is initially at rest until a constant force of F=30N is applied tangentially to the rim of the disk. I-disk==.5MR^2 i. What is the magnitude v of the tangential velocity at a point on the rim after the disk has turned .200 rev? ii. What is the tangential acceleration of a point on the rim of the disk?

Explanation / Answer

Torque t = F*r = 35*0.22= 7.7 Nm Torque t = I a, a is the angular acceleration I for disc about the given axis = mr²/2 = 43.9*0.22²/2 = 1.06238 kg.m² a = 7.7/ 1.06238 = 7.25 rad/s^2 ------------------------------- ?f = v2* a*? Tangential velocity v = r ?f = rv (2* a*?), ? in radian and 120 revolution = 2p120 radian = 0.22v (2*7.25 *2*p*120) v =23 m/s ======================= Tangential acceleration a (T) = r a =0.22*7.25 = 1.595 m/s² Radial acceleration a(r) = v²/r = 23²/0.22 =2404.55 m/s² Resultant acceleration = v[( r a) ²+ (v²/r) ²] = 2404m/s²

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