# Rotational Kinematics - angle between force and velocity

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1. Nov 7, 2015

### trinkleb

Here is the problem I am working on. I have found answers for all of them except part (f), which is the one I need help with. I will report the answers I have so far:

A classic 1957 Chevrolet Corvette of mass 1240 kg starts from rest and speeds up with a constant tangential acceleration of 2.00 m/s^2 on a circular test track of radius 60.0 m. Treat the car as a particle.

(a) What is its angular acceleration? --> 0.0333 rad/s2
(b) What is its angular speed 6.00 s after it starts? --> 0.2 rad/s
(c) What is its radial acceleration at this time? --> 2.4 rad/s2
(d) (I'll skip this one, since it's just a sketching problem)
(e) What are the magnitudes of the total acceleration and net force for the car at this time? --> atot = 3.12 m/s2 and ΣF = 3874 N

(f) What angle do the total acceleration and net force make with the car's velocity at this time?

I'm wondering if I should use one of the rotational kinematics equations, but I'm still not sure how to go about it. Any ideas would be helpful. Thank you!

2. Nov 7, 2015

### K S Thurm

Use your kinematics equations and substitute the variables. v=omega, alpha=a or acceleration, time=time, delta(x)=delta(theta).

Then solve as though kinematics. Use correct units: rad/s rad/s^2 instead of m/s etc. If you need more help, let me know.

3. Nov 7, 2015

### trinkleb

This is what I did using a kinematics equation:

Θ = Θo + ωozt + 1/2(αzt2

Since it starts from rest,

Θ = 0.594 rad = 107°

This is not the right answer. Does Θ really represent the angle between the acceleration and the velocity? I thought it was just the angular displacement, in which I can't see a correlation between that and the acceleration and velocity angle.

4. Nov 7, 2015

### K S Thurm

Delta Theta is Angular displacement. Starts from rest wi=0. Time=6.00s. Alpha=0.0333 rad/s^2. You now have 3 variables. Solve for wf.

Use wf=wi+alpha(time) -----------> wf=0 + (0.0333 rad/s^2)(6.00sec) That should result in your angular speed of 0.1998 or 0.2rad/s

Foe reg kinematics Vf=Vi+at

5. Nov 7, 2015

### K S Thurm

Oh sorry, for part (f) try

Delta Theta = ((wi+wf)/2 ))(time) wi+wf, divide by 2, multiply by time

6. Nov 7, 2015

### trinkleb

Alright, so when I use that equation I get:

ΔΘ = ½(ωo + ωf)(t)
ΔΘ = ½(0 + 0.2 rad/s)(6 s) = 0.6 rad = 34.4°

The book says it should be 50.2°, which doesn't make sense to me. Is there a certain concept that I might be missing which could be keeping me from getting the right answer?

7. Nov 7, 2015

### K S Thurm

You need to use Arctan(ar/at). Try looking up Linear Acceleration in rigid body rotation in your text book. atan=dv/dt = (r)(dw/dt)=ra

Tan^-1(2.40/2.00) =50.19 deg

8. Nov 8, 2015

### trinkleb

Oh wow, that makes a lot of sense. Thank you so much!!