# Homework Help: Rolling without slipping & linear acceleration vector

1. Feb 14, 2016

### Joa Boaz

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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 s

The above was my attempt. Am I on the right track? Or am I doing this wrong?

2. Feb 14, 2016

### haruspex

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.

3. Feb 14, 2016

### Joa Boaz

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

4. Feb 14, 2016

### haruspex

Close, but I think you have some sign errors. It's a bit confusing because if omega is positive then v0 is negative, etc.

5. Feb 14, 2016

### Joa Boaz

Thank

I am sorry but why would a positive omega means a negative VO ?

6. Feb 14, 2016

### haruspex

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.