# Non-Uniform Circular Motion

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.

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

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

Spinnor
Gold Member
Is this a rolling wheel? If so isn't the tangential speed of a point on a moving wheel dependent on the point on the wheel? For the point of contact doesn't the tangential speed go to zero?

Simon Bridge
Homework Helper
@Spinnor: doesn't that depend on your frame of reference?

@BScFTW: you need a bit of context don't you - if the actual paper consistently names the different accelerations then that would be a good guess.

To finesse it, you may want to do it both ways (for when the centripetal alone is 1g and when the total acceleration is 1g) and explain what you are doing.

Personally I'd have done the kinematics in the angular quantities but that way works too.

Spinnor
Gold Member
@Spinnor: doesn't that depend on your frame of reference?

...

Yes, in the frame at rest with respect to the ground the point of contact of the wheel has zero velocity.

Simon Bridge