A flywheel 500mm in diameter is brought uniformly from rest to a speed of 300 rp

Question

A flywheel 500mm in diameter is brought uniformly from rest to a speed of 300 rpm in 20s. Find the velocity and acceleration of a point on the rim 2 seconds after starting from rest.

i have to use these formulas

1.) s = Vot+1/2at^2
2.) V = Vo +at
3.) V^2 = Vo^2 + 2as

s is displacement
Vo is final velocity
V is initial velocity
a = constant acceleration
t = time

           OR

                             

                 

Rotational theta = theta0 + omega0t + 1/2alphat2 omega = omega0 +alphat omega2 = omega +2alpha theta Vt = r omega s=r theta at = ralpha an=v2/r or an = omega2 r

Explanation / Answer

r = radius   =0.5/2 = 0.25 m

w = angular velocity =200rpm = 200*2*3.14/60   =31.4 rad/s

V = velocity = r*w =7.85 m/s

angular acceleartion = W/time = 31.4/20 =1.57 rad/s

at = tangential acceleration = r*angular acceleartion   =0.39 m/s^2

ar = radial acceleration = V^2/r = 246.5 m/s^2

A = total acceleration =[at^2+ar^2]^0.5 =246.49 m/s^2

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