A 60-cm-diameter wheel accelerates uniformly, about a central axle, from 120 rpm

Question

A 60-cm-diameter wheel accelerates uniformly, about a central axle, from 120 rpm to 280 rpm in 4.0 s. Calculate (a) it's angular acceleration, (b) the radial and tangential components of the linear acceleration of a point on the rim of the wheel, 2.0 after it started accelerating and (c) the number of revolutions the wheel turned during the 4-s-acceleration. A 60-cm-diameter wheel accelerates uniformly, about a central axle, from 120 rpm to 280 rpm in 4.0 s. Calculate (a) it's angular acceleration, (b) the radial and tangential components of the linear acceleration of a point on the rim of the wheel, 2.0 after it started accelerating and (c) the number of revolutions the wheel turned during the 4-s-acceleration.

Explanation / Answer

Given:-

Diameter,d=60cm

Radius,r=30cm=0.30m

wi=120 rpm =12.56 rad/s                                   [1rpm=2*pie/60 rad/s]

wf=280 rpm = 29.31 rad/s

t= 4s

(a)

Use the kinematic equation, wf=wi+?t

we get ?= [wf-wi]/t = [29.31-12.56] / 4= 4.19 rad/s2

(b)

Radial component of acceleration , ar= v2/r =rw2

At t= 2s,

w=wi+?t=12.56+[4.19*2] = 20.93 rad/s

ar= rw2 = 0.30* 20.932 = 131.46 m/s2

Radial component of acceleration is 131.46 m/s2

Tangential component of acceleration, at=angular acceleration*radius

at= ?*r= 4.19* 0.30 = 1.26m/s2

Tangential component of acceleration is 1.26 m/s2

(c)

Use the kinematic equation, wf2=wi2+2??

we get ?=[ wf2-wi2] / 2? = [ 29.312-12.562] / 2*4.19= 83.69 rad= 13.33 revolutions [1revolution=2*pie radians]

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