what is the maxium speed in which a 1050 kg car can turn a turn on flat with a 70m radius, when the coefficient of friction is 0.8?

We can use Newton's second law , where Newton force , N = m x g , where m , mass = 1050 kg , and free fall acceleration g = 9.80 m/s2 , since vertical force is getting nullified between car and surface of the road if we make a free body diagram , but only horizontal net force is applicable , so F net = F centripetal when turning with a radius r = 70 meters where coefficient of friction is 0.8 ,

as mv2 / r = coefficient of friction x N ,

so , m v2/ r = coefficient of friction * m * g , so Maximum velocity , v = ( coefficient of friction *r *g )^1/2 ,

Therefore maximum velocity v, = ( 0.8 x 70 m x 9.80 m/sec2 ) ^1/2 = 23.42648074 m/ second

static frictional force Ffr provides the c.f required at maximum speed

Ffr = μN = 0.8(1050)9.81 = 8240.4 N

Ffr = centripetal force = mv²/r = 1050v² / 70 = 8240.4

=> v² = 8240.4 x 70 / 1050 = 23.44²

=> v = 23.44 m/s = 23.44 x 18/5 = 84.38 km/h or 52.43 mph [maximum speed]

[general formula is v(max) = √[μrg], μ is the coefficient of static friction ]

hope this helps

When the outward force exceeds the force of friction, the car will spin out:

mV²/R = µ*m*g

Divide both sides by m

V²/R = µ*g

V² = R*µ*g

V = √R*µ*g] = √70*0.8*9.8] = √548.8] = 23.4m/s

23.4m/s * 1ft/0.3048m * 1mi/5280ft * 3600s/hr = 52.4mi/hr

23.4m/s * 1km/1000m * 3600s/hr = 84.3km/hr

m*ac ≤ m*g*μ

m cross

since ac = V^2/r, then :

V ≤ √g*μ*r ≤ √9.806*0.8*70 ≤ 23.434 m/sec ≤ 84.36 km/h ≤ 52.42 mph

m v^2 / r = mg * coefficient

v^2 / r = g * 0.8

v = sqrt( gr * 0.8) = sqrt( 9.8 * 70 * 0.8) = 23 m/s -> 84 km/hr approx