Answer
i) The object still has centripetal acceleration.
ii) There is tangential acceleration when the angular speed changes.
a) $\omega=8.2\times10^{-2}rad/s$
b) $a=2.62m/s^2$. The direction of acceleration is toward the axis of rotation.
c) The tangential acceleration $a_T$ has a magnitude of $3.22m/s^2$ and is directed tangent to the circular turn and opposite from $\vec{v_T}$.
Work Step by Step
i) An object traveling along a circular path always has centripetal acceleration, so even though the car's tangential speed is constant, it still has centripetal acceleration.
ii) We have $a_T=r\alpha$
So when the car's angular speed changes, which means it has an angular acceleration $\alpha$, it follows that there is now a tangential acceleration $a_T$, too.
a) We have $v_T=32m/s$ and $r=390m$. The car's angular speed is $$\omega=\frac{v_T}{r}=8.2\times10^{-2}rad/s$$
b) As mentioned in i), the car now only has $a_c$, so $$a=a_c=r\omega^2=2.62m/s^2$$
The direction of $\vec{a_c}$ is toward the axis of rotation.
c) We have initial angular speed $\omega_0=8.2\times10^{-2}rad/s$, final angular speed $\omega=4.9\times10^{-2}rad/s$ and time $t=4s$. The car's angular deceleration is $$\alpha=\frac{\omega-\omega_0}{t}=-8.25\times10^{-3}rad/s^2$$
The tangential acceleration is $a_T=r\alpha=-3.22m/s^2$
- Magnitude: $3.22m/s^2$
- Direction: tangent to the circular turn and opposite from $\vec{v_T}$
