Please, I need help with this problem:

A car of mass m travels at a constant speed (v) when it encounters a bump with a radius of curvature r.

➤ Determine, based on the parameters of the problem, the maximum value of v that allows the car to remain adherent to the road at the highest point of the bump. (Result: v= sqrt(gr) )

► Which of the following graphs represents the maximum value of v as a function of r? (Look at the picture)

► How the graph would vary if the car was on the Moon, where g = 1.6m / (s^2)

Thank you in advance

Please I need help with this problem A car of mass m travels at a constant speed v when it encounters a bump with a radius of curvature r Determine based on the class=

Respuesta :

Xaioo

Final-Answer:

To determine the maximum value of v that allows the car to remain adherent to the road at the highest point of the bump, we can consider the forces acting on the car. At the highest point of the bump, the car has two forces acting on it: the gravitational force pulling it downward and the normal force exerted by the road surface. To remain adherent to the road, the centrifugal force must not exceed the maximum frictional force between the tires and the road.

The maximum centrifugal force occurs when the car is at the top of the bump. At this point, the centrifugal force is provided by the equation F = mv²/r, where m is the mass of the car, v is the velocity, and r is the radius of curvature of the bump.

The gravitational force at the top of the bump is given by F = mg, where g is the acceleration due to gravity.

To find the maximum speed v at which the car can traverse the bump without losing contact with the road, the centripetal force required to keep the car moving in a circular path at the top of the bump should be provided by the force of static friction. Equating the centrifugal force to the maximum static frictional force (μN = μmg, where N is the normal force and μ is the coefficient of static friction) gives:

mv²/r = μmg

Here, the mass m cancels out, and rearranging the equation gives v² = μgr. Solving for v, we get v = √(μgr).

The graph depicting the maximum value of v as a function of r should be a curve rising from the origin in a concave-upward manner. This is because as the radius of curvature r increases, the maximum allowable speed v also increases, but the relationship is not linear. The graph should show a nonlinear increase in v as r increases, eventually rising more steeply as r increases.

If the car were on the Moon where the acceleration due to gravity is 1.6 m/s^2, the graph of the maximum value of v as a function of r would change. Since the value of g is smaller on the Moon, the corresponding maximum speed v for a given radius r would also be lower. This change would result in a downward shift of the entire graph, such that for any given value of r, the maximum allowable speed v would now be lower than it was on Earth. The overall shape of the graph, however, with v increasing nonlinearly as r increases, would remain unchanged.

Explanation:

At the top of the hill or bump, there are two forces on the car:

Weight mg pulling down,

Normal force N pushing up.

For the car to just barely stay on the road, N = 0.

Sum the forces in the centripetal direction (towards the center, in this case down):

∑F = ma

mg = m v²/r

v = √(gr)

The velocity is proportional to the square root of the radius, so the graph will look like the second option.

If g were to decrease from 9.8 m/s² to 1.6 m/s², the graph would become flatter.

Ver imagen MathPhys