Beer and Johnston Dynamics 9th 11.139 angular acceleration

[jaredogden]

Homework Statement

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.

Homework Equations

V_r = dr/dt
V_θ = r*dθ/dt
A_r = d²θ/dt²
A_θ = r*d²θ/dt² + 2 dr/dt*dθ/dt
V = r*dθ/dt e_θ
A = -r*(dθ/dt)²e_r + r*d²θ/dte_θ
A_n = v²/ρ
A_t = dv/dt

The Attempt at a Solution

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

A_n1 = (14 ft./s)²/(210 ft.)
A_n1 = 0.933 ft./s²

A_n2 = (24 ft./s)²/(210 ft.)
A_n2 = 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 A_t = Δv/Δt. If I could also find the speed at 2 s, use A_n = 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(A_n/A_t).

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.

 

[gneill]

Angular velocities and accelerations are given in rad/sec and rad/sec² respectively. Presumably your A_n1 and A_n2 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.

 

[jaredogden]

I'm confused when you say presumably your A_n1 and A_n2 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?

 

[gneill]

I'm confused when you say presumably your A_n1 and A_n2 are meant to be angular velocities. I thought those were the normal components of acceleration. The units work out in ft./s² as well.

My apologies. I didn't see the 'squares' (must be going 'terminal blind'). Yes, you've got the centripetal accelerations there.

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?

Yes, and correct. Use the same equations but with angular measures.

 

[jaredogden]

Got it! From v² = v²_o + 2a(x - x_o)

Substituting into get (24 ft./s)² = (14 ft./s)² + 2A_t (95 ft. - 0 ft.)

solving for A_t we get A_t = 2 ft./s²

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

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

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

To find total A take the magnitude of A_t and A_n
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.

 

[gneill]

Well done!