A bullet is fired with speed $100$ m/s from a gun at a point on a circular disc, which is rotating at a constant $0.1$ rad/s, at a distance of $100$ m from the center of the disc. The gun is aiming towards the center of the disc at the moment the bullet is fired. Find the velocity of the bullet as observed in an inertial frame.
I am a little unsure as to what the velocity vector will be for the problem as observed in the rotating frame. Can anyone help?
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The gun is aiming towards the center of the disc and its tangential velocity, resulting from the disc rotation, is perpendicular to that direction. So the resulting velocity of the projectile, as observed in the externa, inertial frame, is a vector sum of two orthogonal velocities: the radial velocity $100\,m/s$ and the velocity in rotation, which is $\omega\cdot r = 0.1\,rad/s \cdot 100\,m = 10\,m/s$.
In the rotating frame the initial velocity of the bullet is of course the velocity given to it by the gun: $100\,m/s$ towards the disc's center. Later, however, the centrifugal force changes the bullet's speed and the Coriolis effect bends its trajectory, so the description of the bullet's movement in the non-inertial frame becomes complicated, a bit...
If you want vectors, write vectors:
Let suppose that, at the moment of the firing, the axes of the two frames coincide and $y=z=0$ for the bullet's position.
$$\mathbf V=(-V_0,0,0)\;;\omega=(0,0,\omega_0)\;;\mathbf Q=(Q_0,0,0)$$
$$\mathbf v = \mathbf{V} + \mathbf\omega \times \mathbf Q =(-V_0,0,0)+(0,0,\omega_0)\times(Q_0,0,0)$$
The bullet is in the the plane of rotation and with some impulse towards the positive $y$ irection (we chose the system to rotate counterclockwise), as expected:
$$\mathbf v=(-V_0,Q_0\omega_0,0)$$ with magnitude $$v=\sqrt{V_0^2+Q_0^2\omega_0^2}$$
Now $V_0=100\,ms^{-1}\;;Q_0=100\,m$ and $\omega_0=0.1\,s^{-1}$, substitute and it's done.
PS: The bullet miss the center of the frame.