Help with centripetal force and friction question please

[takelight2]

Homework Statement

A race car enters a flat 200-m radius curve at a speed of 20.0 m/s while increasing its speed at a constant 2.00 m/s2. If the coefficient of static friction is 0.700, what will the speed of the car be when the car begins to slide?

a- 31.5 m/s
b- 24.3 m/s
c- 28.7 m/s
d- 36.2 m/s
e- 37.1 m/s

Relevant Equations

fc = mv^2/r
Ff = kN

Ff = Fc
(mv^2)/r = kmg
(v^2)/r = kg
v^2 = kgr
v^2 = 0.7*9.8*200
v = 37.04 m/s

I chose option e, and its wrong. What am i doing wrong here?

 

[Chestermiller]

Homework Statement:: A race car enters a flat 200-m radius curve at a speed of 20.0 m/s while increasing its speed at a constant 2.00 m/s2. If the coefficient of static friction is 0.700, what will the speed of the car be when the car begins to slide?

a- 31.5 m/s
b- 24.3 m/s
c- 28.7 m/s
d- 36.2 m/s
e- 37.1 m/s
Relevant Equations:: fc = mv^2/r
Ff = kN

Ff = Fc
(mv^2)/r = kmg
(v^2)/r = kg
v^2 = kgr
v^2 = 0.7*9.8*200
v = 37.04 m/s

I chose option e, and its wrong. What am i doing wrong here?

There is a tangential component of acceleration. What is the magnitude of the acceleration vector?

 

[takelight2]

There is a tangential component of acceleration. What is the magnitude of the acceleration vector?

Tangential acceleration or centripetal acceleration, its a = v^2/r. So would be, 37.04^2/200 = 6.86 m/s^2. How does that help though?

 

[kuruman]

Tangential acceleration or centripetal acceleration, its a = v^2/r. So would be, 37.04^2/200 = 6.86 m/s^2. How does that help though?

Tangential acceleration is different from centripetal acceleration. Only the centripetal acceleration is given by v²/r. What you are doing wrong here is that you are ignoring the tangential component of the acceleration as @Chestermiller remarked.

 

[Lnewqban]

Tangential acceleration or centripetal acceleration, its a = v^2/r.

Tangential means in this case, the effort that each of the driver tires make rearwards, in order to increase the forward velocity of the car.
That force consumes some of the available “lateral traction” of the contact patch of that tire.

Your response would be correct for a tire that is rolling at constant forward velocity
If the driver is accelerating hard, adding significant torque to those driver tires, they will have diminished capability to stand lateral forces of hard cornering.
Therefore, the car will slide sideways at a lower velocity than that of option e).

 

[Chestermiller]

Tangential acceleration or centripetal acceleration, its a = v^2/r. So would be, 37.04^2/200 = 6.86 m/s^2. How does that help though?

$v^2/r$ is the radial (centripetal) component of acceleration. The problem statement says there is also a tangential component of acceleration (i.e., in the direction tangent to the circular path, normal to the radial direction) of 2.00 m/s^2. What is the resultant acceleration?