I really appreciate the help.Finding Force in Nonuniform Circular Motion

[phyzz]

We are back at the very beginning. The problem says that the magnitude of velocity is a linear function of time. Let's denote this f(t) - not v because v is used as a constant specification of initial velocity, so f(0) = v. Tangential acceleration is the time derivative of velocity, i.e., f'(t). What is it?

s = /
u = v
v = 2v
a = ?
t = t

v = u + at
v = at
a = v/t

f'(t) = v/ t
1/v dv = 1/t dt
ln(v) = ln(t) + C
t = 0 when v = 0
I get infinite C?

 

[voko]

In #19 and #25 you wrote what f(t) and f'(t) are. You don't need to solve any differential equations. You just need to determine the constants involved, that's all there is to it.

 

[phyzz]

In #19 and #25 you wrote what f(t) and f'(t) are. You don't need to solve any differential equations. You just need to determine the constants involved, that's all there is to it.

I don't know what you mean

What about the period of one rotation? You said it had to be included somewhere

 

[voko]

There is no period. The motion is not uniform.

I do not think this is going in the right direction. I do not feel that you understand the significance of linearity of velocity. I will make one final attempt.

The magnitude of the velocity is a linear function of time. That means $v(t) = a_{\tau}t + v_0$. The derivative is $v'(t) = a_{\tau}$. $a_{\tau}$ is tangential acceleration, $v_0$ is the initial velocity, which is denoted just as $v$ in the problem, which is very unfortunate because it got you badly confused earlier. You have already deduced $a_{\tau}$ in terms of $R$ and $v_0$ (again, it was confusingly denoted simply as $v$). With that, you can easily restore $v(t)$ and thus obtain $a_n$, the normal (or radial) acceleration. Then it is trivial to get the magnitude of net acceleration and force.

 

[phyzz]

There is no period. The motion is not uniform.

I do not think this is going in the right direction. I do not feel that you understand the significance of linearity of velocity. I will make one final attempt.

The magnitude of the velocity is a linear function of time. That means $v(t) = a_{\tau}t + v_0$. The derivative is $v'(t) = a_{\tau}$. $a_{\tau}$ is tangential acceleration, $v_0$ is the initial velocity, which is denoted just as $v$ in the problem, which is very unfortunate because it got you badly confused earlier. You have already deduced $a_{\tau}$ in terms of $R$ and $v_0$ (again, it was confusingly denoted simply as $v$). With that, you can easily restore $v(t)$ and thus obtain $a_n$, the normal (or radial) acceleration. Then it is trivial to get the magnitude of net acceleration and force.

Thank you for your patience. So the radial accn would be aR = v(t)^2 / R

v(t) being what was found earlier from $v(t) = a_{\tan}t + v_0$