At a time t= 3.40 s , a point on the rim of a wheel with a radius of 0.250 m has

Question

At a time t= 3.40 s , a point on the rim of a wheel with a radius of 0.250 m has a tangential speed of 52.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.2 m/s^2 . Calculate the wheel's constant angular acceleration. Alpha= rad/s^2 Calculate the angular velocity at t= 3.40 s . omega 0= rad/s Calculate the angular velocity at t= 0. omega 0= rad/s Through what angle did the wheel turn between t = 0 and t= 3.40 s ? theta= rad At what time will the radial acceleration equal g = 9.81 m/s^2 ? t= s

Explanation / Answer

a.constant angular acceleration=tangential acceleration/radius

=10.2/.25= - 40.8 rad/sec^2

b.angular velocity = v/r
= 208 at t=3.40

c.we know the angular acceleration so can add that on to the velocity that we have at t=3.40.

w(final)=w(initial)+.5 Angular acceleration*time

=346.72rad/s at t=0

d.

s=ut+0.5at^2

its the same for angular movement

theta = 346.72*3.4 - .5 * 40.8 * 3.4*3.4

=943./024 rad

e.

mrw^2 = F * r

dividing my mass on both sides:

rw^2=g

w=w0+at, where w0 is at t=3.4
t=5.09

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