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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 F = ma size 12{F= ital "ma"} {} . For uniform circular motion, the acceleration is the centripetal acceleration— a = a c size 12{a=a rSub { size 8{c} } } {} . Thus, the magnitude of centripetal force F c size 12{F rSub { size 8{c} } } {} is

F c = m a c . size 12{F rSub { size 8{c} } =ma rSub { size 8{c} } } {}

By using the expressions for centripetal acceleration a c size 12{a rSub { size 8{c} } } {} from a c = v 2 r ; a c = 2 size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } ;``a rSub { size 8{c} } =rω rSup { size 8{2} } } {} , we get two expressions for the centripetal force F c size 12{F rSub { size 8{c} } } {} in terms of mass, velocity, angular velocity, and radius of curvature:

F c = m v 2 r ; F c = mr ω 2 . size 12{F rSub { size 8{c} } =m { {v rSup { size 8{2} } } over {r} } ;``F rSub { size 8{c} } = ital "mr"ω rSup { size 8{2} } } {}

You may use whichever expression for centripetal force is more convenient. Centripetal force F c size 12{F rSub { size 8{c} } } {} is always perpendicular to the path and pointing to the center of curvature, because a c size 12{a rSub { size 8{c} } } {} is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for r size 12{r} {} , you get

r = mv 2 F c . size 12{r= { { ital "mv" rSup { size 8{2} } } over {F rSub { size 8{c} } } } } {}

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

The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.
The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the F c size 12{F rSub { size 8{c} } } {} , the smaller the radius of curvature r size 12{r} {} and the sharper the curve. The second curve has the same v size 12{v} {} , but a larger F c size 12{F rSub { size 8{c} } } {} produces a smaller r size 12{ { {r}} sup { ' }} {} .

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 c = mv 2 r . Thus,

F c = mv 2 r = ( 900 kg ) ( 25.0 m/s ) 2 ( 500 m ) = 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 μ s N size 12{μ rSub { size 8{s} } N} {} , where μ s size 12{μ rSub { size 8{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 = mg . Thus the centripetal force in this situation is

F c = f = μ s N = μ s mg . size 12{F rSub { size 8{c} } =f=μ rSub { size 8{s} } N=μ rSub { size 8{s} } ital "mg"} {}

Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for F c size 12{F rSub { size 8{c} } } {} from the equation

F c = m v 2 r F c = mr ω 2 } , size 12{ left none matrix { F rSub { size 8{c} } =m { {v rSup { size 8{2} } } over {r} } {} ##F rSub { size 8{c} } = ital "mr"ω rSup { size 8{2} } } right rbrace ,} {}

Questions & Answers

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this is the mass of an atom of an element in ratio with the mass of carbon-atom
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its s.i unit is Nm
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Force×perpendicular distance N×m=Nm
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pressure
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how do you calculate the 5% uncertainty of 4cm?
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4cm/100×5= 0.2cm
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= 200g±(5%)10g
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the relationship between the applied force and the deflection
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Practice Key Terms 5

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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