# 6.3 Centripetal force

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• 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

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

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Miranwa
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Toheeb
principle is a rule or law of nature, or the basic idea on how the laws of nature are applied.
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principle is a rule or law of nature, or the basic idea on how the laws of nature are applied.
tathir
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why does weight change but not mass?
Theo
Theo, the mass of an object can change but it depends on how you define that object. First, you need to know that mass is the amount of matter an object has, and weight is mass*gravity (the "force" that attracts object A to the object B mass).
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So if you face object A with object B, you will get a different result than facing object A with object C, so the weight of object A changes but not its mass.
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Now, if you have an object and you take a part away from it, you are changing it mass. Lets use the human body and fat loss process as an example.
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When you lose weight by doing exercise, you are being attracted by the same object before and after losing weight so the change of weight is related to a change of mass not a change of gravity.
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Theo, weight =mass. gravity, here mass is fixed everywhere but gravity change in different places so weight change not mass.
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Yusuf Shuaibu, for water the Adhessive force between water molecules and glass is greater than the cohessive force between it's own molecules but for Mercury the cohessive force will be greater in comparison with adhessive force. For this water wet glass but Mercury does not.
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wieght is the vector
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Yes
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what is specific heat capacity of watee
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@siyanbola Resistance to acceleration
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Micheal
lawrence
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rate of change of velocity is acceleration
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its amount of heat to raise the temlrature through one kelvin of substance .
ghulam
The amount of heat energy required to raise the température of water by 1K
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infact a it must be a unit mass of water
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approximately equal to 4184J/Kg/K
Cffrrcvccgg
Just got through thermodynamics last semester. Also a change in 1 degree in celcius is equivalent to a change in 1 degree kelvin
Dillon
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Dillon
I think, at least
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Numericals 🙄
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4200kJ/kgk...
Trevor
J
Trevor
SHM and uniform circular motion
Ishaq