Elson

How bout torque = force times radius = ma times radius ->mra times radius. torque = 2I times angular acc? For gear A, 0.75= 2(0.02) times angular acc. Therefore i get angular acc = 0.75/0.04 = 18.75?

Well i used torque = I(moment of inertia) x a(angular acceleration) 0.75 = (1/2)(4)(0.01) x a, a = 0.75/0.02, a=37.5 but the answer is 5.4rad/s2. As for the tangential contact forces, i have no idea about that. I was informed late by the University and missed the class for 2 weeks. Need help from you guys.. thanks!

Hello, does anyone know how to do this question? Gear A (radius = 150mm, mass= 4kg and radius of gyration = 100mm) and gear B(radius = 450mm, mass= 10kg and radius of gyration= 300mm) rotate about their respective centrers. A constant clockwise torque of 0.75Nm is applied to the gear A. neglecting bearing friction torque, determine the angular acceleration of each gear and the tangential contact forces between the gears.

Ahha!!.. Thanks for the help!!!

Hello i'm new to this forum and i'm facing some problems while dealing with this question. A particle is moving along the x axis in SHM. It starts from the equilibrium position and moves towards the right at t = 0 s. The amplitude of the motion is 2 cm, and the frequency is 1.5 Hz. 1)The equation for velocity can be determined by differentiating the function for displacement w.r.t. t, V = AWn cos (Wn t) Therefore, the maximum speed will be |V max| = A(Wn)(1) <---- what is this? where does the (1) come from? 2)The time that it first occurs could be found as follow; cos(Wnt) = −1, <--- why is it -1? Wt = cos−1(−1), t = 0.33 s.