Question

A 220-kg speedboat is negotiating a circular turn (radius = 31 m) around a buoy. During the turn, the engine causes a net tangential force of magnitude 590 N to be applied to the boat. The initial tangential speed of the boat going into the turn is 9.9 m/s. (a) Find the tangential acceleration. (b) After the boat is 6.0 s into the turn, find the centripetal acceleration.

Answer 1

Refer to the figure shown below, which illustrates the problem.

The applied tangential force of 590 N  creates an applied torque of
T = (590 N)*(31 m) = 18290 N-m

The rotational moment of inertia of the boat is
I = (220 kg)*(31 m)² = 211420 kg-m²

Part a.
The angular acceleration is
α = T/I = (18290 N-m)/(211420 kg-m²) = 0.0865 rad/s²
The tangential acceleration is
$a_{t}$ = (0.0865 rad/s²)*(31 m) = 2.68 m/s²

Part b.
The initial tangential speed is v = 9.9 m/s, therefore the initial angular speed is
ω₀ = (9.9 m/s)/(31 m) = 0.3194 rad/s
After 6 s, the angular speed is
ω = (0.3194 rad/s) + (0.0865 rad/s²)*(6 s) = 0.8384 rad/s

The centripetal acceleration is
$a_{n}$ = (31 m)*(0.8384 rad/s)² = 21.79 m/s²