A marked point on a 60cm in diameter wheel has a tangential speed (V(tan)) of 3.00 m/s when it begins to slow down with a tangential acceleration (A(tan)) of -1.00 m/s^2.(adsbygoogle = window.adsbygoogle || []).push({});

a) what are the magnitudes of the points angular velocity (ω) and acceleration(α) at t=1.5 seconds?

b)At what time is the magnitude of the points acceleration equal to g?

I mostly need help with b...

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For tangential, I will put the variable with (ta) next to it. For Radial (ra).

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a) starting with angular acceleration (which I assume must be constant) α=A(ta)/r

A(ta) = -1.00 and r= 0.30m, so α=-3.33 rad/s^2

For Angular Velocity (ω), ω=V(ta)/r, but we need to find V(ta) at 1.5 seconds first.

V(final)=V(initial) + A(ta)Δt

=3.00+(-1.00)(1.5)

=1.50 m/s

Therefore ω=1.5/0.30 = 5 rad/s

Done part a)

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b) No idea. I know that there are three accelerations to deal with (tangential, radial and the vector), and that you must set the vector one to -9.8.... I also understand that A(vector)^2=A(ta)^2+A(radial)^2.... at least I think thats where i need to go with it...

any help is great help!!!

Thanks

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# Homework Help: Non-Uniform Circular Motion

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