# Beer and Johnston Dynamics 9th 11.139 angular acceleration

1. Oct 8, 2011

### jaredogden

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

An outdoor track is 420 ft. in diameter. A runner increases her speed at a constant rate from 14 to 24 ft./s over a distance of 95 ft. Determine the total acceleration of the runner 2 s after she begins to increase her speed.

2. Relevant equations

Vr = dr/dt
Vθ = r*dθ/dt
Ar = d2θ/dt2
Aθ = r*d2θ/dt2 + 2 dr/dt*dθ/dt
V = r*dθ/dt eθ
A = -r*(dθ/dt)2er + r*d2θ/dteθ
An = v2
At = dv/dt

3. The attempt at a solution

diameter = 420 ft. therefore ρ = .5*420 ft. or ρ = 210 ft.

An1 = (14 ft./s)2/(210 ft.)
An1 = 0.933 ft./s2

An2 = (24 ft./s)2/(210 ft.)
An2 = 2.74 ft./s2

I am not sure where to go from here. I know I can't use equations from rectilinear motion since this is angular. If I could find the time it takes her to run the 95 ft. I think I could use that to find an average tangential acceleration by At = Δv/Δt. If I could also find the speed at 2 s, use An = v2/ρ and find the normal component of acceleration. Taking the magnitude of the two would give me the total acceleration and α = tan-1(An/At).

I'm just not sure of the next step. I'm also not sure if that is the right approach.. Thanks for your time and any help ahead of time.

Last edited: Oct 8, 2011
2. Oct 8, 2011

### Staff: Mentor

Angular velocities and accelerations are given in rad/sec and rad/sec2 respectively. Presumably your An1 and An2 are meant to be angular velocities.

You can 'co-opt' the equations from linear motion if you change all the variables to their angular equivalents. So for example, v = a*t becomes ω = $\alpha$*t. You can probably think of a linear equation that relates initial velocity, final velocity, acceleration, and distance.

3. Oct 8, 2011

### jaredogden

I'm confused when you say presumably your An1 and An2 are meant to be angular velocities. I thought those were the normal components of acceleration. The units work out in ft./s2 as well.

Also, you are saying that I can use the equations for uniformly accelerated rectilinear motion for angular acceleration? I would assume that if I were able to use them they would give me the tangential component of acceleration correct?

4. Oct 8, 2011

### Staff: Mentor

My apologies. I didn't see the 'squares' (must be going 'terminal blind'). Yes, you've got the centripetal accelerations there.
Yes, and correct. Use the same equations but with angular measures.

5. Oct 8, 2011

### jaredogden

Got it! From v2 = v2o + 2a(x - xo)

Substituting in to get (24 ft./s)2 = (14 ft./s)2 + 2At (95 ft. - 0 ft.)

solving for At we get At = 2 ft./s2

Using v = vo + at substituting in for vo = 14 ft./s a = At = 2 ft./s2 and the given t = 2 s

v = 14 ft./s + 2 ft./s2*(2 s)
v = 18 ft./s

Using An = v2/ρ substituting in the v we just found and ρ = 210 ft.
An = (18 ft./s2)/210 ft.
An = 1.54 ft./s2

To find total A take the magnitude of At and An
A = sqrt((2ft./s2)2 + (1.54 ft./s2)2)
A = 2.53 ft./s2

EDIT: And thanks a ton for your helping along. I didn't think I could use those equations since the equations for finding the other values since the equations for finding velocity and acceleration changed for angular. However now it makes sense why I can.

6. Oct 8, 2011

Well done!