Rolling without slipping & linear acceleration vector

AI Thread Summary
The discussion revolves around the physics of rolling without slipping, focusing on deriving linear acceleration vector equations for various points on a wheel. Key calculations include determining the vehicle's speed and vector accelerations based on given parameters such as radius, angular velocity, and angular acceleration. The participants engage in clarifying the relationships between the vectors and the signs of angular quantities, emphasizing the importance of direction in vector equations. There is a consensus on the need for careful differentiation of vector expressions to accurately describe motion. The conversation highlights the complexities of understanding rolling motion and the significance of coordinate systems in these calculations.
Joa Boaz
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Homework Statement


Rolling without slipping

A) Derive the linear acceleration vector equations for points A, B, C, and O in terms of R, ω, α and θ at this instant.

B) R = 0.5 m, ω=-54 r/s and α = 0. Determine the MPH of the vehicle and the vector accelerations of points A, B, C, and O.

C) R = 0.5 m, ω=-54 r/s and α = +4.9 rad/sec/sec. Find the magnitude and direction of the acceleration of points O & C.

D) R = 0.5 m, ω=-54 r/s and α = +4.9 rad/sec/sec. Determine the magnitude and direction of the acceleration of the vehicle in "g" units and how long it would take to stop.

media-f53-f537a8aa-70c1-4eab-86cd-d6b687139ba8-phpuy2kkq.png


Homework Equations

The Attempt at a Solution


A)
AA = R α j + Rc (-ω2 R) i
AB = 2 R α i + ω2 R j
AO = R α i
AC = ω2 R j

B)
V = R ω
V = 0.5 x 54
= 27 m/s
= 60.4 mph

AA = -ω2 R = -1458 i m/s2
AB = - 1458 j
AO = 0
AC = 1458 j

C)
At O AO = R α i ⇒ 0.5 x 4.9 = 2.45 i m/s2
At C AC = 1458 j

D)
A = R α
= 2.45 m/s2

A = 0.25 g

t= ω/α
t= 11.02 sThe above was my attempt. Am I on the right track? Or am I doing this wrong?
 
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I disagree with several of your answers in part A. (What is RC?) Haven't looked beyond that.
I think the easiest way is to write a vector expression for the location of a point on the circumference in terms of the location of the centre and an angle theta (i.e. theta =0 would be point A, etc.) then differentiate as necessary.
 
Thank you.

So, if i wrote a general velocity vector equation for any point on the periphery of the wheel

Vp = Vo + ω X Rop

By differentiating so
AA = AC + ω × (ω × RAC) + α × RAC
 
Joa Boaz said:
Thank you.

So, if i wrote a general velocity vector equation for any point on the periphery of the wheel

Vp = Vo + ω X Rop

By differentiating so
AA = AC + ω × (ω × RAC) + α × RAC
Close, but I think you have some sign errors. It's a bit confusing because if omega is positive then v0 is negative, etc.
 
haruspex said:
Close, but I think you have some sign errors. It's a bit confusing because if omega is positive then v0 is negative, etc.

Thank

I am sorry but why would a positive omega means a negative VO ?
 
Joa Boaz said:
Thank

I am sorry but why would a positive omega means a negative VO ?
According to the diagram, theta is measured anticlockwise, so a positive omega would mean rolling to the left. The x-axis is positive to the right.
 

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