A car approaches the top of a hill that is shaped like a vertical circle with a radius of \(55.0 \mathrm{m} .\) What is the fastest speed that the car can go over the hill without losing contact with the ground?

Short Answer

Expert verified
Answer: The maximum speed the car can go over the hill without losing contact with the ground is approximately 26.1 m/s.

Step by step solution

01

Problem Data

We have a hill shaped like a vertical circle with a radius of 55.0 m. Let's denote the speed of the car as v.
02

Centripetal Force and Gravitational Force

The centripetal force acting on a body moving in a circle is \(F_c = \dfrac{mv^2}{r}\), where m is the mass of the car, v is its speed, and r is the radius of the circle (55.0 m in this case). The gravitational force acting on the car is \(F_g = mg\), where g is the acceleration due to gravity (approximately \(9.8 \,\text{m/s}^2\)).
03

Critical Condition for Loss of Contact

The car will lose contact with the ground when the centripetal force is equal to the gravitational force, as they both act on the same object. The normal force acting between the car's tires and the ground decreases to zero when the car loses contact. Thus, we can write the equation for the critical condition as: \(F_c = F_g\) or \(\dfrac{mv^2}{r} = mg\).
04

Mass Cancellation and Calculating Speed

Notice that the mass m appears on both sides of the equation, so we can cancel it out: \(\dfrac{v^2}{r} = g\). Now we just need to solve for v. Rearrange the equation to find v: \(v^2 = rg\) and then take the square root of both sides to get: \(v = \sqrt{rg}\).
05

Plugging in Values and Solving for V

Now we can plug in the known values for r and g: \(v = \sqrt{(55.0 \, \text{m})(9.8 \, \text{m/s}^2)}\). Calculate the value for v to get the fastest speed the car can go over the hill without losing contact with the ground: \(v \approx 26.1 \, \text{m/s}\). Thus, the fastest speed that the car can go over the hill without losing contact with the ground is \(\boxed{26.1\,\text{m/s}}\).

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