I need Physics help!!?

Question

A car of mass of 1500kg traveling at 45.0km/hr enters a banked turn covered with ice. The road is banked at an angle , and there is no friction between the road and the car's tires. What is the radius of the turn if the angle = 20.0 deg(assuming the car continues in uniform circular motion around the turn)?

Answer 1

Since there is no friction in this problem, the only force keeping the car from flying off is due to the weight of the car.

Drawing a free-body diagram, you see that only the cosine portion of the weight directly affects the centripetal force.

Therefore,

m*g*cos(theta) = m*v^2/r. The masses cancel in the equation, now solve for r:

r = v^2/(g*cos(theta)). Remember, convert your velocity into SI units (m/s). Thus:

(45 km/hr)(1000 m / 1 km)(1 hr / 3600 sec) = 12.5 m/s

Plugging in our known values, solve for r:

r = (12.5)^2/(9.8*cos(20)) -----------> r = 16.97 meters

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Hope this helps

Answer 2

Given: m=1500kg
v=45km/hr
20 degree=angle of bank

Fc=mv^2/r, where Fc is the centripetal force in N, m is the mass of the car in kg, v is the speed of the car in m/s, and r is the radius of the turn in m.

First, convert v=45km/hr to m/s
=45*1000/3600
=12.5m/s

Note: there are 1000m in a km, and 3600s in an hr.

Draw a force diagram showing weight of car and its component along the banked road. We are given its mass=1500kg. We convert this to weight by multiplying it by g=9.8. Hence weight=1500*9.8=14700N. This force is
pointing downward. Its component along the banked road is 14700sin20=14700*.342=5027N. This is our Fc or centripetal force.

Now substitute known values in the formula:

Fc=mv^2/r
5027=1500*12.5^2/r
r=1500*12.5^2/5027
=46.6m

Answer 3

Sum of the forces = 0.
This means that the cosine of the downward force (mg) must equal the centripital force of the car.