# 6.3 Centripetal force

 Page 1 / 10
• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.

Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force    . The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net $\text{F}=\text{ma}$ . For uniform circular motion, the acceleration is the centripetal acceleration— $a={a}_{c}$ . Thus, the magnitude of centripetal force ${\text{F}}_{\text{c}}$ is

${\text{F}}_{\text{c}}={m\text{a}}_{\text{c}}.$

By using the expressions for centripetal acceleration ${a}_{c}$ from ${a}_{c}=\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{a}_{c}={\mathrm{r\omega }}^{2}$ , we get two expressions for the centripetal force ${\text{F}}_{\text{c}}$ in terms of mass, velocity, angular velocity, and radius of curvature:

${F}_{c}=m\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{F}_{c}=\text{mr}{\omega }^{2}.$

You may use whichever expression for centripetal force is more convenient. Centripetal force ${F}_{\text{c}}$ is always perpendicular to the path and pointing to the center of curvature, because ${\mathbf{a}}_{c}$ is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for $r$ , you get

$r=\frac{{\mathrm{mv}}^{2}}{{F}_{c}}\text{.}$

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

## What coefficient of friction do car tires need on a flat curve?

(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.

(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see [link] ).

Strategy and Solution for (a)

We know that ${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}$ . Thus,

${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}=\frac{\left(\text{900 kg}\right)\left(\text{25.0 m/s}{\right)}^{\text{2}}}{\left(\text{500 m}\right)}=\text{1125 N.}$

Strategy for (b)

[link] shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is ${\mu }_{\text{s}}N$ , where ${\mu }_{\text{s}}$ is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that $N=\mathit{mg}$ . Thus the centripetal force in this situation is

${F}_{\text{c}}=f={\mu }_{\text{s}}N={\mu }_{\text{s}}\text{mg}\text{.}$

Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for ${F}_{\text{c}}$ from the equation

$\begin{array}{c}{F}_{\text{c}}=m\frac{{v}^{2}}{r}\\ {F}_{\text{c}}=\text{mr}{\omega }^{2}\end{array}\right\},$

a15kg powerexerted by the foresafter 3second
what is displacement
movement in a direction
Jason
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
Hi
saeid
hi
Yimam
What is thê principle behind movement of thê taps control
what is atomic mass
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Explain why it is difficult to have an ideal machine in real life situations.
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
can i get all formulas of physics
yes
haider
what affects fluid
pressure
Oluwakayode
Dimension for force MLT-2
what is the dimensions of Force?
how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
haider
topic of question?
haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
Hello group...
Chioma
hi
haider
well hello there
sean
hi
Noks
hii
Chibueze
10g
Olokuntoye
0.2m
Olokuntoye
hi guys
thomas
the meaning of phrase in physics
is the meaning of phrase in physics
Chovwe