chris and john are driving around in circles.chris notices that it takes 4.7 secs to make a complete revolution.he also notices his car is just on the verge of sliding sideways. if coeffcient of friction=1.07 for rubber on asphalt ,and the car(and passengers) have a mass of 1500kg, what is the radius of the turn?
the answer is 5.87m
2007-03-29 02:14:20 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
math is hard head hurts got to stop now
2007-03-29 02:30:52 · answer #1 · answered by mobile auto repair (mr fix it) 7 · 0⤊ 1⤋
It is the frictional force (F) that is providing the centripetal force.
F = normal force * coefficient of friction
F = m * g * 1.07 Newtons
Centripetal force = m * w^2 * r
where w is the angular velocity in rads/sec r is the radius in metres.
Therefore
m * g * 1.07 = m * w^2 * r
r = g * 1.07/ w^2 (It doesn't depend on m at all!!)
1 revolution = 2 * pi radians
w = 2 * pi / 4.7 radians /sec = 1.337 rad / sec
r = 9.81 * 1.07 / 1.337^2 = 5.873 metres
2007-03-29 09:47:59 · answer #2 · answered by lunchtime_browser 7 · 0⤊ 0⤋
Circular motion requires that a centripetal force must be acting on the car.In the given situation, friction provides the necessary centripetal force.
Centripetal force F =mass 'm'*velocity'v'^2 / radius 'r'
As velocity 'v'=circumference / time period 'T'
therefore Centripetal force F =mass 'm'*radius'r'*4pie square / time period Tsquare
Radius 'r'=(T/2pie)^2*F / m
Friction =co efficient of friction*m*acceleration due to gravity 'g'
As friction provides the centripetal force F , equating friction with F we get,
radius 'r'=(T/2pie)^2*co efficient of friction*g
substituting T=4.7 s, co efficient of friction=1.07, 'g'=9.8m/s^2and pie=3.14 we get
radius 'r'=5.8685=5.87m
2007-03-29 10:17:01 · answer #3 · answered by ukmudgal 6 · 0⤊ 0⤋