A cyclist is rounding a 20m radius curve at 12 m/s. What is the minimum possible…

the concepts required to solve this problem are newton’s second law and centripetal acceleration.

first use the newton’s second law to solve for the expression of minimum coefficient of static friction required.

You're reading: A cyclist is rounding a 20m radius curve at 12 m/s. What is the minimum possible…

finally, substitute the values the in the coefficient equation to calculate the minimum possible coefficient of static friction.

the newton’s second law states
who the net force on an object is the product of mass of the object and final acceleration of the object. the expression of newton’s second law is,

f=ma\sum f = ma

here,

f\sum f

is the sum of all the forces on the object,

mm

is mass of the object, and

aa

is the acceleration of the object.

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the expression of static friction over a horizontal surface is,

fμsmgf \le \mu _\rmsmg

here,

μs\mu _\rms

is the coefficient of static friction,

mm

is mass of the object, and

gg

is the acceleration due to gravity.

the centripetal acceleration is the rate of change of the velocity in tangential direction. the magnitude of acceleration is constant for a uniform circular motion. it is always towards the center along the radius vector of any circular motion. the centripetal acceleration is expressed as,

ac=v2ra_\rmc = \fracv^2r

here,

vv

is the constant speed, and

rr

is the perpendicular distance from the object to center of rotation.

use the newton’s second law for the bike.

substitute

ff

for

f\sum f

in the equation

f=ma\sum f = ma

.

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substitute

μsmg\mu _\rmsmg

for

ff

in the equation

f=maf = ma

.

μsmg=mv2rμs=v2rg\beginarrayc\\\mu _\rmsmg = m\fracv^2r\\\\\mu _\rms = \fracv^2rg\\\endarray

use the equation of coefficient of static friction.

substitute

12m/s12\rm m/s

for

vv

,

20m20\rm m

for

rr

, and

9.81m/s29.81\rm m/\rms^2

for

gg

in the equation

μs=v2rg\mu _\rms = \fracv^2rg

and calculate the

μs\mu _\rms

.

μs=(12m/s)2(20m)(9.8m/s2)=0.735\beginarrayc\\\mu _\rms = \frac\left( \rm12 m/s \right)^2\left( 20\rm m \right)\left( 9.8\rm m/\rms^2 \right)\\\\ = 0.735\\\endarray

the minimum coefficient of static friction between the bike tires and the ground is

0.7350.735

.

Source: https://tonupboys.com
Category: Bike Tires