Rotational Kinematics of a Particle

[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

 

[Doc Al]

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

Why cosine? (And why change to radians?)

 

[Destrio]

will making a triangle give me any information?
doing the same thing with tan and sin don't 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

 

[Doc Al]

will making a triangle give me any information?

Sure.

doing the same thing with tan and sin don't give me good answers

Why not? One of them will.

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

Yes, but that's the same triangle.

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

 

[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 don't know what I was thinking initially

but this makes more sense than what i was doing

thanks!