Rotation about a Fixed Axis

[Vladimir_Kitanov]

Homework Statement

On picture

Relevant Equations

$\theta = \theta_0 + \omega _0 t + \frac {1}{2} \alpha t^2$
$\omega = \omega_0 + \alpha t$
$\omega^2 = \omega_0^2 + 2 \alpha (\theta - \theta_0)$
$a_t = \alpha * r$
$a_n = \omega^2 * r$

Answer should be (c) 32,7 $\frac{m}{s^2}$

My attempt:

$\omega_{2\pi}$ -> $\omega$ after 1 revolution
$\omega_{2\pi} = 0,2 * (2\pi)^2$
$\omega_{2\pi} = 7,9 \frac{rad}{s}$

$\frac {d}{dt}\omega = \alpha = 0,2*2*\theta$
$\alpha_{2\pi}$ -> $\alpha$ after 1 revolution
$\alpha_{2\pi} = 2,51 \frac {rad}{s}$

$a_n$ -> normal acceleration
$a_t$ -> tangential acceleration

$a_t = \alpha_{2\pi} * r = 1,26 \frac {m}{s^2}$
$a_n = \omega^2 * r = 32,205 \frac {m}{s^2}$
$a = \sqrt{a_n^2 + a_t^2} = 31,2 \frac{m}{s^2}$

 

[erobz]

$\frac {d}{dt}\omega = \alpha = 0,2*2*\theta$
$\alpha_{2\pi}$ -> $\alpha$ after 1 revolution
$\alpha_{2\pi} = 2,51 \frac {rad}{s}$

Does $\alpha$ have the proper units of angular acceleration in your derivation?

 

[jbriggs444]

$\frac {d}{dt}\omega = \alpha = 0,2*2*\theta$

I am concerned with:
$\frac {d}{dt}\omega = \alpha = 0,2*2*\theta$

You are given $\omega$ as a function of $\theta$. But you appear to have differentiated with respect to $\theta$ and obtained a derivative with respect to $t$.

 

[malawi_glenn]

This might be easier to think about the differentiation
$\omega = 0.2 \cdot (\theta(t))^2$

Note, some of the formulas you listed as "relevant" does not apply here. Why is it so?

 

[Vladimir_Kitanov]

Note, some of the formulas you listed as "relevant" does not apply here. Why is it so?

I just put them there.
They are useful in some problems like that.

 

[Vladimir_Kitanov]

Does $\alpha$ have the proper units of angular acceleration in your derivation?

No $\alpha$ have unit $\frac{rad}{s^2}$
But that does not change anything.

 

[Vladimir_Kitanov]

I am concerned with:
$\frac {d}{dt}\omega = \alpha = 0,2*2*\theta$

You are given $\omega$ as a function of $\theta$. But you appear to have differentiated with respect to $\theta$ and obtained a derivative with respect to $t$.

That is my mistake.
Thanks.

It should be:
$\alpha = 0,2*2*\theta*\omega$
Derivative of $\theta$ with respect to time is $\omega$

 

[erobz]

No $\alpha$ have unit $\frac{rad}{s^2}$
But that does not change anything.

But it does.

For future reference: If you derive something, and its units do not match the quantity you are after then there is a mistake in said derivation.

 

[kuruman]

To summarize, $$\alpha=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\omega \frac{d\omega}{d\theta}.$$What is the tangential component $a_t$?

 

[Vladimir_Kitanov]

$$\alpha=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\omega \frac{d\omega}{d\theta}.$$What is the tangential component $a_t$?

$a_t = \alpha * r$
$\alpha$ is angular acceleration.

 

[erobz]

The term “linear” acceleration is confusing in the problem statement. What I think is being calculated is just the magnitude of the acceleration?

 

[Vladimir_Kitanov]

The term “linear” acceleration is confusing in the problem statement. What I think is being calculated is just the magnitude of the acceleration?

Yes

 

[bob012345]

Yes

You got the basic concept correct in your OP but just are having issues with the numerical calculations. It is the vector sum as you tried to compute originally. You just need to be very careful computing $a_n$ and $a_t$ especially the part explained by @kuruman in post #9.

 

[Steve4Physics]

$a_t = \alpha_{2\pi} * r = 1,26 \frac {m}{s^2}$
$a_n = \omega^2 * r = 32,205 \frac {m}{s^2}$
$a = \sqrt{a_n^2 + a_t^2} = 31,2 \frac{m}{s^2}$

In addition to the other comments, can I add that the hypotenuse is the longest side of a right-triangle.

(Minor edit.)