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.

 

[CellCoree]

I would first clarify that the problem states the point is moving with constant tangent acceleration. This means that the point is moving along the half circle at a constant speed, but its direction is constantly changing. This is different from constant normal acceleration, where the speed and direction of the point would be changing.

With this clarification, we can approach part (c) by considering the components of the acceleration vector. Since the point is moving along the half circle at a constant speed, the tangential component of the acceleration must be zero. This means that the entire acceleration vector is pointing towards the center of the circle, and its magnitude is equal to the normal component of the acceleration.

Using this information, we can calculate the modulus of the mean vector of the total acceleration:

$$| \langle {\mathbf w} \rangle | = \sqrt{({\mathbf w}_t)^2 + ({\mathbf w}_n)^2} = \sqrt{0^2 + ({\mathbf w}_n)^2} = |{\mathbf w}_n| = \frac{\langle v \rangle ^2}{R} = \frac{(\pi R)^2}{\tau^2} = \frac{2 \pi R}{\tau^2} \approx 10 \frac{\text{cm}}{\text{s}^2}.$$

This matches the given answer and confirms that the point is indeed moving with a constant normal acceleration of approximately 10 cm/s^2.