A 63-cm-diameter wheel accelerates uniformly about its center from 130 rpm to 25

Question

A 63-cm-diameter wheel accelerates uniformly about its center from 130 rpm to 250 rpm in 4.0 s.

A) Determine its angular acceleration.

B) Determine the radial component of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating. Don't forget that the wheel has an initial angular velocity when it started accelerating.

C) Determine the tangential component of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating

Explanation / Answer

Here,

radius ,r = 63/2 = 31.5 cm

w1 = 130 rpm = 13.611 rad/s

w2 = 250 rpm = 26.2 rad/s

time , t = 4 s

a) let the angular accelearation is a

a = (w2 - w1)/t

a = (26.2 - 13.611)/4

a = 3.141 rad/s^2

tha angular acceleration is 3.141 rad/s^2

B)

radial component of linear acceleartion = a * radius

radial component of linear acceleartion = 3.141 * 0.315

radial component of linear acceleartion = 0.989 m/s

C)

let the angular velcity after 2 s is wf

wf = w1 + a * t

wf = 13.611 + 3.141 * 2

wf = 19.9 rad/s

tangential component of acceleration = wf^2 * r

tangential component of acceleration = 19.9^2 * 0.315

tangential component of acceleration = 124.7 m/s^2

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