At a time t = 2.90 s , a point on the rim of a wheel with a radius of 0.190 m ha

Question

At a time t = 2.90 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 10.5 m/s2 .

Calculate the wheel's constant angular acceleration.

Calculate the angular velocity at t = 2.90 s .

Calculate the angular velocity at t=0.

Through what angle did the wheel turn between t=0 and t = 2.90 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

=a/r

=-10.5/0.190
-55.26 rad/s^2

=v/r
=47.0/0.190
247.3 rad/s

=0*t+.5**t^2
47/r=0-10.5*2.9/r

solve fore 0

0=47/r+10.5*2.9/r
0=407.63

=0*t+.5**t^2
885.6 rad

rad acc=^2*r=9.81
=sqrt(9.81/0.19)
51.6315 rad/s

=0+*t
t=(-0)/

t=(51.6315-407.63)/-55.26
6.44 s

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