Question

A body of mass 2 kg moves in a (counterclockwise) circular path of radius 3 meters, making one revolution every 5 seconds. You may assume the circle is in the xy-plane, and so you may ignore the third component.

A. Compute the centripetal force acting on the body.
B. Compute the magnitude of that force.

Answer 1

a)
`\vec{F_(c)}=mR\omega^(2)`

`\vec{F_(c)}=9.53 \: N \vec{r}`

b)
`|\vec{F_(c)}|=9.53 \: N`

Explanation:

a) The centripetal force equation is:

`\vec{F_(c)}=m\vec{a_(c)}`

`\vec{F_(c)}=mR\omega^(2)`

Now, we know that the body makes one revolution every 5 seconds, so we can find the angular velocity:

`\omega=(1 rev)/(5 s)=0.2\: (rev)/(s)=1.26\: (rad)/(s)`

`\vec{F_(c)}=2*3*1.26^(2)=9.53 \: N \vec{r}`

The centripetal force is a vector in the radius direction.

b) The magnitude of that force will be:

`|\vec{F_(c)}|=9.53 \: N`