a pulley with rotational inertia of 1.5 times 10 ^-3 kg times m^2 about its axle and a radius of 15 cm is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F=.5t + .3 t^2, where F is in newtons and t in seconds. The pulley is initiall at rest. At t=2s what is the angualr veloctiy?

Question

Christiaan · Answer

First we calculate the torque.

t=rxF=r.F.sin(b) where is the angle between r and F (which in this case equals 90°).
So in this case t= r.F = 0,15m.F = 0.075t + 0.045t²

We also know that I = 1.5 . 10^-3 kg.m²
and so we can calculate the angular acceleration (a) with the formula:

t = I.a
=> a = t/I = 50t + 30t²

Now, we need to find the angular velocity after 2 seconds. Since a=dw/dt (with w, the angular velocity), we can find the angular by taking the definite integral of a for t from 0 to 2

=> w = integral of 50t+30t² from 0 to 2=
(25.(2)²+10(2)^3)-(25.(0)²+10(0)^3)= 180-0 = 180

so the angular velocity after two seconds = 180 rad/s

ed · Answer

thank you soooooooo much i really appreciate you taking your time to help