A car starts from rest and accelerates around a flat curve of radius R = 36 m. T

Question

A car starts from rest and accelerates around a flat curve of radius R = 36 m. The tangential component of the car's acceleration remains constant at a_tangential = 3.3 m/s^2, while the centripetal acceleration increases to keep the car on the curve as long as possible. The coefficient of friction between the tires and the road is 0.95. What distance does the car travel around the curve before it begins to skid? (Be sure to include both the tangential and centripetal components of the acceleration).

Explanation / Answer

Inital angular velocity of the car is 0 = 0.0 rad /s Inital angular velocity of the car is 0 = 0.0 rad /s Tangential accelaration of the car   at = 3.3 m/ s2 The radius of the circle    r = 36 m The coeffitient of frtiction   = 0.95 Under circular motion                  centripetal force = mg                                  mv2 / r = mg    Linear velocity                   v = sqrt ( gr ) Substiting all values in the above equation we get                                           v = sqrt ( 0.95 * ( 9.8 m/s2 )( 36 m )                                           v = 18.3 m/s Final angular velocity of the car is                                            = v /r                                                = ( 18.3 m/s / 36 m)                                                = 0.508 rad /s ------------------------------------------------------- The tangential accelartion   at = r angular accelaration              = at /r                                               = ( 3.3 m/s2 )   / 36 m                                               = 0.09167 rad /s2 According to rotational kinematic equations                                          2 =  02   + 2 Here , angular distance tavelled by the car is                                            = ( 2 - 02 ) / 2 Substitute all values in the above equation , we get                                              = ( 0.508 rad /s )2 / (2 * 0.09167 rad /s2 )                                               = 2.77 rad The distance travelled by the car on the curve before it skidding is                                             s = r *                                                = ( 36 m ) ( 2.77 rad )                                                  = 99.72 m According to rotational kinematic equations                                          2 =  02   + 2 Here , angular distance tavelled by the car is                                            = ( 2 - 02 ) / 2 Substitute all values in the above equation , we get                                              = ( 0.508 rad /s )2 / (2 * 0.09167 rad /s2 )                                               = 2.77 rad The distance travelled by the car on the curve before it skidding is                                             s = r *                                                = ( 36 m ) ( 2.77 rad )                                                  = 99.72 m

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