Q.A rider of mass $40\ \text{kg}$ sits at the rim of a merry-go-round of radius $4\ \text{m}$ rotating at a constant angular speed. In the rotating frame attached to the merry-go-round, the rider experiences an apparent (centrifugal) outward force of $200\ \text{N}$. Find the angular speed of rotation and the linear (tangential) speed of the rider.

Given: mass $m = 40\ \text{kg}$, radius $r = 4\ \text{m}$, apparent (centrifugal) force in the rotating frame $F = 200\ \text{N}$.\n\nIn the rotating frame attached to the merry-go-round, the centrifugal pseudo-force has magnitude $F = m\omega^2 r$. Solving for $\omega$: $\omega^2 = \frac{F}{mr} = \frac{200}{40 \times 4} = \frac{200}{160} = 1.25 \Longrightarrow \omega = \sqrt{1.25} \approx 1.12\ \text{rad/s}$\n\nThe rider's linear (tangential) speed, as seen from the ground frame, is then $v = \omega r = 1.12 \times 4 \approx 4.47\ \text{m/s}$\n\n(This is also, necessarily, exactly the speed that would be found by instead equating the real centripetal force in the ground frame, $mv^2/r$, to the same $200\ \tex …

Set the given apparent force equal to $m\omega^2 r$ and solve for $\omega$, then find …

• Forgetting to take the square root after solving for $\omega^2$, and reporting $1.25$ as the angular speed itself. …