Finding centripetal acceleration for a car on a curved road

[MYS9]

If there is a car with a mass of 1200 kg and the radius of the curvature is 25m, and the speed is 43km/h, what is the acceleration?

How do I find the acceleration

At what speed would the centripetal force equal the force of gravity? ( I got 15.7 m/s but I'm not sure if this is right)
If the car was going faster than the speed calculated above, what would happen to the car?

Thanks!

 

[ShawnD]

If the car is going at a constant speed on a round track with a given radius, the acceleration is

$$a = \frac{V^2}{r}$$

v is instantaneous velocity (speed) and r is radius.



I got the same velocity as you (15.66 m/s)

 

[M98Ranger]

To find the acceleration, we can use the formula a = v^2/r, where v is the velocity in m/s and r is the radius in meters. First, we need to convert the given speed of 43km/h to m/s by dividing it by 3.6. This gives us a speed of 11.94 m/s. Plugging this into the formula along with the given radius of 25m, we get an acceleration of approximately 0.46 m/s^2.

To find the speed at which the centripetal force equals the force of gravity, we need to set the two forces equal to each other. The centripetal force is given by Fc = mv^2/r, where m is the mass of the car, v is the velocity, and r is the radius. The force of gravity is given by Fg = mg, where m is the mass of the car and g is the acceleration due to gravity (9.8 m/s^2). Setting these two equations equal to each other and solving for v, we get v = √(rg). Plugging in the given mass of 1200 kg, we get a speed of approximately 15.7 m/s, which is the same result you got.

If the car was going faster than this calculated speed, the centripetal force would be greater than the force of gravity, causing the car to experience a net inward force. This would result in the car turning inwards towards the center of the curve, potentially causing it to lose control or skid off the road. This is why it is important to always drive at safe speeds and follow the speed limit on curved roads. I hope this helps!