Problem 114 - University Physics with Modern Physics

Question

A 70 -kg person rides in a 30 -kg cart moving at 12 \(\mathrm{m} / \mathrm{s}\) at the top of a hill that is in the shape of an arc of a circle with a radius of 40 \(\mathrm{m} .\) (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart may travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

Step-by-Step Solution

Answer

(a) 434 N; (b) 19.8 m/s, independent of mass.

Step 1: Understand the Problem

The problem provides the mass of a person, a cart, their speed, and the radius of a circular path. We need to find the apparent weight of the person over the top of the hill and the maximum speed without losing contact with the surface.

Step 2: Determine Forces Acting at the Top of the Hill

As the cart passes over the top of the hill, two forces act on the person: gravitational force \( F_g = mg \) directed downward and centripetal force requirement \( F_c = \frac{mv^2}{r} \), also downward. The apparent weight of the person is the normal force \( N \), which satisfies \( mg - N = \frac{mv^2}{r} \).

Step 3: Calculate Apparent Weight (a)

Substitute values into the equation from the previous step: \( m = 70 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( v = 12 \text{ m/s} \), \( r = 40 \text{ m} \). Solve for \( N \): \[ N = mg - \frac{mv^2}{r} = 70 \times 9.8 - \frac{70 \times 12^2}{40} \].

Step 4: Solve for Normal Force (a)

Calculate the equation: \( N = 70 \times 9.8 - \frac{70 \times 144}{40} = 686 - 252 = 434 \text{ N} \). The apparent weight of the person is 434 N.

Step 5: Find Maximum Speed for No Contact Loss (b)

For no contact, the normal force \( N \) should be zero. This means the gravitational force is entirely providing the necessary centripetal force: \( mg = \frac{mv^2}{r} \). Solve for \( v \).

Step 6: Calculate Maximum Speed (b)

Cancel out \( m \) and solve: \( v^2 = gr \Longrightarrow v = \sqrt{gr} = \sqrt{9.8 \times 40} \). Calculate \( v \): \( v = \sqrt{392} \approx 19.8 \text{ m/s} \).

Step 7: Discuss Mass Dependency (b)

The maximum speed does not depend on the mass of the cart or the person, as the mass \( m \) cancels out when solving \( mg = \frac{mv^2}{r} \). Therefore, only gravitational acceleration and the radius affect the maximum speed.

Key Concepts

Centripetal Force

When moving in a circular path, objects experience a force that pulls them towards the center of the circle. This is known as centripetal force. It's not a force in itself but rather the result of several force components acting together, like tension, gravity, or friction. For our problem, as the person and cart move at the top of the hill, this centripetal force is directed downward. It's computed using the formula

• \( F_c = \frac{mv^2}{r} \)

where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the circular path. Remember, centripetal force is necessary to keep the object moving in a circle. If this force weren't present, the object would travel in a straight line instead of a circular path. At the top of the hill, the centripetal force requirement is satisfied primarily by gravity.

Gravitational Force

Gravitational force is the attraction between two masses. Near Earth's surface, this is the force pulling us towards the ground, commonly referred to as weight. The gravitational force acting on an object can be calculated using the equation:

• \( F_g = mg \)

where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \). In our scenario, this force acts downward on the person and the cart as they go over the hill. It provides the necessary force to maintain circular motion and partially determines the apparent weight, which is the force felt by the person during this motion.

Normal Force

The normal force is the support force exerted by a surface, perpendicular to the surface, to ensure that an object stays on a path. It balances the forces acting on an object to prevent it from sinking into the surface or floating away. In circular motions, like at the top of the hill, the normal force is what we perceive as the apparent weight. At the hilltop, we calculate it as

• \( N = mg - \frac{mv^2}{r} \)

This equation tells us that the normal force is less than the gravitational force due to the requirement of providing the centripetal force. If the speed were high enough for this force to reach zero, the cart would lose contact with the surface, causing a moment of weightlessness.

Circular Motion

Objects in circular motion move along a path that forms a circle at a constant speed. This does not mean constant velocity, as velocity is a vector dependent on both speed and direction. In circular motion, even if speed remains constant, direction constantly changes. For stability in such motion, especially at the apex of a curve like the hill in our problem, specific conditions must be met:

• The centripetal force requirements must be supported by gravitational and normal forces.
• Circular motion requires continuous inward force (centripetal force) to prevent it from moving off its path.

As velocity increases, the necessary centripetal force increases, too, until the point where gravity alone can't keep the cart on the path, and normal force becomes zero (no contact).

Mass and Weight Relationship

Mass is a measure of how much matter an object contains, whereas weight is the force exerted by the gravitational attraction of the Earth on that mass. The formula that connects both is:

• \( ext{Weight} = mg \)

where weight is the gravitational force acting on the mass \( m \) with gravity \( g \). In the scenario at the top of the hill, an interesting aspect is how mass affects maximum speed. While mass directly determines weight and necessary centripetal force, it does not affect the maximum speed to prevent losing contact because the formula for maximum speed with gravitational force as centripetal force simplifies to

• \( v = \sqrt{gr} \)

This cancels out mass, showing that only gravitational acceleration \( g \) and radius \( r \) influence this maximum speed, not the mass of the person or cart.