Question

What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a spaceship taking a circular turn of radius 4360 km at a speed of 39000 km/h?

Answer 1

Answer:

$(a)w=2.485*10^{-3}m/s\\(b)a=26.92m/s^2\\(c)\alpha =0rad/s^2$

Explanation:

Given data

Linear speed of spaceship V=39000 km/h =10833.3 m/s

The radius of circular motion r=4360 km=4360000 m

For Part (a) angular velocity

The angular velocity is given by:

$w=\frac{v_{speed}}{r_{radius}}$

Substitute the given values

So

$w=\frac{10833.3m/s}{4360000m} \\w=2.485*10^{-3}m/s$

For Part(b) The radial acceleration of spaceship

The radial acceleration is given by:

$a_{r}=w^2 r$

Substitute given values and value of ω

So

$a_{r}=(2.485*10^{-3}m/s)^2(4360000m)\\a_{r}=26.92m/s^2$

For Part (c) The tangential acceleration

The spaceship is rotating with constant angular velocity,then the  tangential acceleration of spaceship is zero

So

$\alpha =0rad/s^2$