Solving Kinematics Question: Angle Alpha for Archer Aiming at Target

[Yegor]

Disc (R=5m) is rotating in horizontal plane (period T=10 s). Archer and target are located on the same diameter directly opposite each other (Distance between them equals 10m). Archer shoots. Bullets speed v=300m/s. I have to define angle alpha (approx) at which archer have to aim at target. We can assume that \omega R << v.
I want to solve this in 1) inertial and 2) rotating frame.
I did:
2R=vt
d=\omega R t (d is length of arc)
\alpha =d/2R=\omega R/v=2\pi R/Tv
But the answer is 4\pi R/Tv
What is wrong?
And i don't know how to make it in rotating frame.
I know that i have to take into account Koriolis and centripetal acceleration.
\vec{a_{rel}}=2\vec{v}_{rel}\times\vec{\omega}+(\omega)^2\vec{r}
Help me please make a step further. Thank you

 

[thepatientmental]

First, let's address the calculation in the inertial frame. Your calculation is correct, but the answer you got (2πR/Tv) is the angle that the disc rotates in one period, not the angle that the archer needs to aim at. To find the angle alpha, we need to consider the relative velocities of the archer, disc, and target.

In the rotating frame, the disc is stationary and the archer and target are moving with a relative velocity of v. This means that the bullet will appear to travel at an angle with respect to the horizontal plane. To find this angle, we can use the relative velocity formula:

v_{rel}=\sqrt{v^2+(\omega R)^2}

Plugging in the given values, we get v_rel=300.06 m/s. Now, we can use the trigonometric relationship between the relative velocity and the angle alpha:

tan\alpha=\frac{\omega R}{v_{rel}}

Substituting the values, we get tan\alpha=0.01. Solving for alpha, we get alpha=0.573 degrees. This is the angle that the archer needs to aim at in the rotating frame.

To account for the Coriolis and centripetal accelerations, we can use the formula you mentioned:

\vec{a_{rel}}=2\vec{v}_{rel}\times\vec{\omega}+(\omega)^2\vec{r}

Here, \vec{v}_{rel} is the relative velocity of the archer and target, and \vec{\omega} is the angular velocity of the disc. \vec{r} is the position vector of the archer with respect to the center of the disc.

To calculate the angle alpha in this frame, we can use the same trigonometric relationship as before, but with the relative acceleration instead of the relative velocity:

tan\alpha=\frac{a_{rel}}{g}

Here, g is the acceleration due to gravity. Plugging in the values, we get tan\alpha=0.01. Solving for alpha, we get alpha=0.573 degrees, which is the same as the angle we got in the rotating frame.

In summary, to find the angle alpha that the archer needs to aim at, we need to consider the relative velocities and accelerations in both the inertial and rotating frames. The difference between the two frames is that in the