# Rotational Kinematics of a Particle

1. Dec 5, 2007

### Destrio

A rigid body, starting at rest, rotates about a fixed axis with a constant angular acceleration α. Consider a particle a distance r from the axis. Express (a) the radial acceleration and (b) the tangential acceleration of this particle in terms of α, r and time t.
c) if the resultant acceleration of the particle at some instant makes an angle of 57.0 degrees with the tangential acceleration, through what total angle has the body rotated from t=0 to that instant.

i got radial acceleration = rω^2 = (α^2)(t^2)r
tangential acceleration = αr

for c i made a triangle and got cos(57deg) = cos(.99rad) = (α^2)(t^2)r/αr = αt^2

so i plugged it into the kinematics formula
theta = (1/2)αt^2
theta = (1/2)cos(.99rad)
theta = .27 radians
but that is incorrect

2. Dec 5, 2007

### Staff: Mentor

Why cosine? (And why change to radians?)

3. Dec 5, 2007

### Destrio

will making a triangle give me any information?
doing the same thing with tan and sin dont give me good answers

is the resultant acceleration = sqrt([(α^2)(t^2)r]^2 + [αr]^2)
should I be making a triangle with that?

thanks

4. Dec 5, 2007

### Staff: Mentor

Sure.
Why not? One of them will.

Yes, but that's the same triangle.

Just draw a right triangle with two sides being the two acceleration components.

5. Dec 5, 2007

### Destrio

tan(57deg) = (α^2)(t^2)r / αr
tan(57deg) = αt^2

theta = (1/2)αt^2
theta = (1/2)tan(57deg)
theta = .7699 rad

.7699rad * 360/2pi = 44.1 degrees

i dont know what I was thinking initially

but this makes more sense than what i was doing

thanks!

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