Circular motion

Since the radial acceleration of the particle (which, together with the table, is constrained to move with angular speed $\omega = 5$ ) is zero, we deduce that $\ddot{r} = r\omega^2 = 25r$ and hence $\dot{r}^2 - 25r^2$ is constant. Thus $\dot{r} = 5\sqrt{r^2 - 25}$ and hence $\dot{r} = 5\times12 = 60$ when $r = 13$ . The particle has a radial speed of $60$ , but it also has a transverse speed of $13\times5 = 65$ , and hence its total speed is $5\sqrt{12^2 + 13^2} = 5\sqrt{313} = \boxed{88.459...}$ ms ${}^{-1}$ .

We will consider a rotating frame of reference, where there is a centripetal force of $F=m\omega^2 r$ acting on the particle. Here $\omega=5 rad/s$ is the angular speed, $m$ is the mass and $r$ is the distance from the center. The equation of motion is $ma=m\omega^2 r$ or

$\ddot r = \omega^2 r$

We look for the solution in the form of $r=A e^{\beta t}+B e^{-\beta t}$ . We get

$A\beta^2 e^{\beta t}+B \beta^2 e^{-\beta t} = \omega^2 (A e^{\beta t}+B e^{-\beta t})$

and this is satisfied if $\beta=\omega$ . The initial conditions are that at $t=0$ we have $r_0=5m$ and $v=\dot r=0$ . Since $\dot r=A \omega e^{\omega t}-B \omega e^{-\omega t}$ we get $A+B=r_0$ and $A-B=0$ . With $A=B=r_0/2$ we can re-write the solution as

$r=r_0 \cosh \omega t$

$\dot r =r_0 \sinh \omega t$

Due to $\cosh^2 a - \sinh^2 a = 1$ we have

$\dot r = \omega \sqrt{r^2-r_0^2}$

When we insert $r=13m$ , this yields $\dot r=60m/s$ .

Added on 4/3/2018, after Mark Hennings's comment:

Seen from the inertial frame of reference the particle also has a tangential velocity of $v=\omega r=65m/s$ . The magnitude of the total velocity is $v=\sqrt{60^2+65^2}=88.5m/s$ .