At what time is the acceleration of the marked point equal to g?

AI Thread Summary
The discussion revolves around calculating the angular velocity and acceleration of a point on a wheel with a diameter of 60 cm, which has a tangential speed of 3.00 m/s and a tangential acceleration of -1.00 m/s². The user successfully determines the angular acceleration to be -3.33 rad/s² and the angular velocity at 1.5 seconds to be 5 rad/s. For determining when the point's total acceleration equals gravitational acceleration (g), the user acknowledges the need to consider tangential, radial, and vector components of acceleration. There is also a debate about the frame of reference affecting the tangential speed of points on the wheel, particularly at the point of contact with the ground. The discussion highlights the complexity of the problem and the importance of context in solving it.
BScFTW
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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

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 that's where i need to go with it...

any help is great help!

Thanks
 
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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?
 
@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.
 
Simon Bridge said:
@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.
 
Spinnor said:
Yes, in the frame at rest with respect to the ground the point of contact of the wheel has zero velocity.
Strictly speaking, a particular point on the circumference of the wheel is stationary when it meets the ground. The point of contact usually moves in any frame fixed to the ground.
But meh.

I wonder how that would make a difference to the problem in OP? I mean - is the motion of the marked point even circular in that frame?
 
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