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

a)
`d=628.3m`

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

c)
`a_(cp)=10m/s^2`

Step-by-step 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=(v^2)/(2d)=((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)=(v^2)/(r)=((10m/s)^2)/(10m) =10m/s^2`