Question

An object moves in a circular orbit starting from rest. The orbit has a radius of 10 m. It takes 10 revolutions to reach its final speed of 10 m/s. a) How far did it travel from rest to its final speed (in meters)?
b) If it accelerated at a constant rate, what was its tangential accelerate during that time?
c) Once it reaches its final speed, what is its centripetal acceleration?

Answer 1

Answer:

a)$d=628.3m$

b)$a_t=5m/s^2$

c)$a_{cp}=10m/s^2$

Explanation:

a) If it did 10 revolutions, it returned to the original point, but it travelled 10 times the circunference $C=2\pi r$, where r is the radius of the circle, so we have $d=10(2\pi r)=20\pi (10m)=628.3m$

b) The tangential component of this problem can be treated as a one dimensional problem, so the (tangential) acceleration $a_t$ needed to reach the final speed given starting from rest after traveling a distance d can be obtained using the equation for accelerated motion $v^2=v_0^2+2a_td=2a_td$, so we have:

$a_t=\frac{v^2}{2d}=\frac{(10m/s)^2}{2(10m)}=5m/s^2$

c) We use our values in the formula for centripetal acceleration given the tangential velocity and radius:

$a_{cp}=\frac{v^2}{r}=\frac{(10m/s)^2}{10m} =10m/s^2$