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 :

whitneytr12

Answer:

a) [tex]\vec{F_{c}}=mR\omega^{2}[/tex]

[tex]\vec{F_{c}}=9.53 \: N \vec{r}[/tex]

b) [tex]|\vec{F_{c}}|=9.53 \: N [/tex]

Step-by-step explanation:

a) The centripetal force equation is:

[tex]\vec{F_{c}}=m\vec{a_{c}}[/tex]

[tex]\vec{F_{c}}=mR\omega^{2}[/tex]

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

[tex]\omega=\frac{1 rev}{5 s}=0.2\: \frac{rev}{s}=1.26\: \frac{rad}{s}[/tex]

[tex]\vec{F_{c}}=2*3*1.26^{2}=9.53 \: N \vec{r}[/tex]

The centripetal force is a vector in the radius direction.

b) The magnitude of that force will be:

[tex]|\vec{F_{c}}|=9.53 \: N [/tex]

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