MHB Kinematics, particle on half circle

[Fantini]

Here's the problem.

A point traversed half a circle of radius $R = 160 \text{ cm}$ during a time interval of $\tau = 10.0 \text{ s}$. Calculate the following quantities averaged over that time:

(a) the mean velocity $\langle v \rangle$;

(b) the modulus of the mean velocity $ |\langle {\mathbf v} \rangle|$;

(c) the modulus of the mean vector of the total acceleration $| \langle {\mathbf w} \rangle |$ if the point moved with constant tangent acceleration.

I'm having trouble with (c). Since he mentioned there is a constant tangent acceleration I assumed it is non-zero (this may not be the case). I did not manage to reach a conclusion. Assuming it is zero, the mean acceleration is merely the mean normal acceleration. This leads to

$$| \langle {\mathbf w} \rangle | = 2 | {\mathbf w}_n | = 2 \frac{\langle v \rangle^2}{R} = 2 \frac{\pi^2 R}{\tau^2}.$$

I'm missing something, because the alleged answer is

$$| \langle {\mathbf w} \rangle | = \frac{2 \pi R}{\tau^2} \approx 10 \frac{\text{cm}}{\text{s}^2}.$$

Thank you. :)

 

[I like Serena]

Here's the problem.

A point traversed half a circle of radius $R = 160 \text{ cm}$ during a time interval of $\tau = 10.0 \text{ s}$. Calculate the following quantities averaged over that time:

(a) the mean velocity $\langle v \rangle$;

(b) the modulus of the mean velocity $ |\langle {\mathbf v} \rangle|$;

(c) the modulus of the mean vector of the total acceleration $| \langle {\mathbf w} \rangle |$ if the point moved with constant tangent acceleration.

I'm having trouble with (c). Since he mentioned there is a constant tangent acceleration I assumed it is non-zero (this may not be the case). I did not manage to reach a conclusion. Assuming it is zero, the mean acceleration is merely the mean normal acceleration. This leads to

$$| \langle {\mathbf w} \rangle | = 2 | {\mathbf w}_n | = 2 \frac{\langle v \rangle^2}{R} = 2 \frac{\pi^2 R}{\tau^2}.$$

I'm missing something, because the alleged answer is

$$| \langle {\mathbf w} \rangle | = \frac{2 \pi R}{\tau^2} \approx 10 \frac{\text{cm}}{\text{s}^2}.$$

Thank you. :)

Hey Fantini! ;)

Let $\alpha$ be the angular acceleration, which is constant.
Let $\omega_0$ be the initial angular velocity.
And let $0 \le \theta \le \pi$.

Then:
$$| \langle {\mathbf w} \rangle |
= \left| \frac{\Delta \mathbf v}{\Delta t} \right|
= \left| \frac{\mathbf v(\pi) - \mathbf v(0)}{\tau} \right|
= \frac{|-R(\alpha\tau + \omega_0) + R\omega_0|}{\tau}
= R\alpha$$

Furthermore, we have:
$$\theta(\tau) = \frac 12 \alpha \tau^2 + \omega_0 \tau = \pi$$
$$\alpha = \frac{\pi - \omega_0\tau}{\frac 12 \tau^2}$$

So:
$$| \langle {\mathbf w} \rangle |
= R\frac{\pi - \omega_0\tau}{\frac 12 \tau^2}
=\frac{2\pi R}{\tau^2} - \frac{2\omega_0 R}{\tau}$$

Apparently your problem assumes that $\omega_0=0$.

 

[Fantini]

Hey ILS! :)

This is problem 1.19 from Irodov's Problems in General Physics, 1988. Is there any loss of generality by assuming $\omega_0 = 0$? It seems not, since the circular movement begins after $t=0$.

Thank you for your insightful input. ;)

 

[I like Serena]

Hey ILS! :)

This is problem 1.19 from Irodov's Problems in General Physics, 1988. Is there any loss of generality by assuming $\omega_0 = 0$? It seems not, since the circular movement begins after $t=0$.

Thank you for your insightful input. ;)

You are right that we can assume $t=0$ without loss of generality.
However, since $|\langle \mathbf w \rangle|$ changes with $\omega_0$, it seems to me that there is loss of generality. So I don't understand where this assumption is coming from.