Non-uniform circular motion and tangential acceleration

[Symstar]

Homework Statement

An object of mass m is constrained to move in a circle of radius r. Its tangential acceleration as a function of time is given by a_{tan} = b + ct^2, where b and c are constants.

A) If v = v_0 at t = 0, determine the tangential component of the force, F_{\tan }, acting on the object at any time t > 0.
Express your answer in terms of the variables m, r, v_0, b, and c.

B) Determine the radial component of the force F_{\rm{R}}.
Express your answer in terms of the variables m, r, v_0, b, t, and c.

Homework Equations

a_{tan} = b + ct^2
a_r=\tfrac{v^2}{r}
Newton's Laws

The Attempt at a Solution

A. was not a problem for me:
F_{\tan}=ma_{\tan}=m(b+ct^2)

For B.:
F_R=ma_r
a_r=\tfrac{v^2}{r}
It seems to make sense that because v is tangential speed we could use...
v(t)=v_0+a_{\tan}t=v_0+(b+ct^2)t
So that...
a_r=\frac{(v_0+(b+ct^2)t)^2}{r}
Finally giving...
F_R=m(\frac{(v_0+(b+ct^2)t)^2}{r}

Which is not correct. What did I do wrong?

 

[tiny-tim]

Welcome to PF!

Hi Symstar! Welcome to PF!

It seems to make sense that because v is tangential speed we could use...
v(t)=v_0+a_{\tan}t

That only works if a_tan is constant, doesn't it?

Hint: dv/dt = … ?

 

[Symstar]

Hi Symstar! Welcome to PF! That only works if a_tan is constant, doesn't it?

Hint: dv/dt = … ?

dv/dt = a_tan correct?

So would I need to integrate?
\int a_{tan} = \int b + ct^2
\frac{dv}{dt}= bt+\tfrac{1}{3}ct^3

And it seems logical in our case that +C would actually be +v₀

Which would end up giving me:
F_R=m\frac{(v_0+bt+\tfrac{1}{3}ct^3)^2}{r}

Which I just confirmed to be the correct answer... thanks for you your help tim.