Question

A body of mass 8 kg moves in the xy-plane in a counterclockwise circular path of radius 7 meters centered at the origin, making one revolution every 5 seconds. At the time t=0, the body is at the rightmost point of the circle. A. Compute the centripetal force acting on the body at time t.

Answer 1

`F_(cp)=88.43N`

Step-by-step explanation:

The equation for centripetal force is given by Newton's 2nd Law:

`F_(cp)=ma_(cp)`

The equation for centripetal acceleration is
`a_(cp)=r\omega^2`

If the object makes one revolution (an angle of
`2\pi rad`) every 5 seconds it means that its angular velocity is
`\omega=(2\pi rad)/(5s)`.

Putting all together, and for our values:

`F_(cp)=ma_(cp)=mr\omega^2=(8kg)(7m)((2\pi rad)/(5s))^2=88.43N`