At a time t = 3.20 s , a point on the rim of a wheel with a radius of 0.190 m ha

Question

At a time t = 3.20 s , a point on the rim of a wheel with a radius of 0.190 m has a tangential speed of 47.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.8 m/s2 . Calculate the wheel's constant angular acceleration. Calculate the angular velocity at t = 3.20 s .Calculate the angular velocity at t=0.Through what angle did the wheel turn between t=0 and t = 3.20 s ?Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81 m/s2 ?

Explanation / Answer

Angular acceleratio = tangentiall accc. / Radius

        = ( 10.8 ) / ( 0.190 ) = 56.84 rad/s^2

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angular velocity = tangential velocity / radius = 47 / 0.190 = 247.37 rad/s

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using wf = wi   + alpha*t

247.37 = wi   + ( -56.84 x 3.20)

wi =429.26 rad/s

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for angle,

wf^2 - wi^2 = 2 x alpha x theta

247.37^2 - 429.26^2    =2 x -56.84 x theta

theta = 1082.60 rad

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radial acc. = w^2 r

9.81 = w^2 x 0.190

w = 7.19 rad/s

using wf - wi   = alpha*t

7.19 - 429.26 =   -56.84 * t

t = 7.43 sec

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