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

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Homework Help Overview

The problem involves a marked point on a wheel with a specified diameter, which is experiencing tangential deceleration. The original poster seeks assistance in determining when the point's acceleration equals the acceleration due to gravity (g). The discussion touches on concepts of angular velocity, tangential and radial acceleration, and the relationship between these quantities.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of angular acceleration and velocity, with some questioning the nature of the wheel's motion and the frame of reference. The original poster expresses uncertainty about how to approach the problem of finding when the point's acceleration equals g, mentioning the need to consider tangential, radial, and vector components of acceleration.

Discussion Status

The discussion is ongoing, with participants providing insights into the problem's context and raising questions about the assumptions made regarding the wheel's motion and the relevant accelerations. Some guidance has been offered regarding the relationship between different types of acceleration, but no consensus has been reached on the approach to solve part b of the problem.

Contextual Notes

There is a mention of the need for clarity regarding the definitions of different accelerations and how they relate to the frame of reference, particularly in the context of a rolling wheel. The original poster's understanding of the problem setup may be incomplete, as indicated by the questions raised by other participants.

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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