Problem 53 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A \(975-\mathrm{kg}\) sports car (including driver) crosses the rounded top of a hill (radius = 88.0 \(\mathrm{m} )\) at 12.0 \(\mathrm{m} / \mathrm{s}\) . Determine \((a)\) the normal force exerted by the road on the car, \((b)\) the normal force exerted by the car on the \(72.0-\mathrm{kg}\) driver, and \((c)\) the car speed at which the normal force on the driver equals zero.

Step-by-Step Solution

Answer

(a) 7973.84 N, (b) 587.96 N, (c) 29.38 m/s speed for zero normal force.

Step 1: Understand the Problem

We need to calculate three different aspects of forces at the top of a hill. These involve finding the normal force on the car and driver using given parameters and finding the speed at which the driver's normal force becomes zero.

Step 2: Identify Forces at the Hilltop

At the top of the hill, the forces acting on the car are gravity and the normal force. The normal force and the gravitational force act in opposite directions, with gravity providing the centripetal force. This can be expressed by the equation: \[ mg - N = \frac{mv^2}{r} \] where \( m = 975 \) kg is the mass of the car, \( g = 9.81 \) m/s\(^2\) is the acceleration due to gravity, \( N \) is the normal force, \( v = 12 \) m/s is the velocity, and \( r = 88 \) m is the radius of the hill.

Step 3: Solve for Part (a) - Normal Force on the Car

Rearrange the equation from Step 2 to solve for the normal force \( N \):\[ N = mg - \frac{mv^2}{r} \] Substitute the given values:\[ N = (975 \times 9.81) - \frac{975 \times (12)^2}{88} \] Calculate:\[ N = 9564.75 - 1590.91 = 7973.84 \text{ N} \] Hence, the normal force exerted by the road on the car is 7973.84 N.

Step 4: Solve for Part (b) - Normal Force on the Driver

The normal force acting on the driver can be determined similarly. Rearrange to solve for the driver's normal force \( N_d \):\[ N_d = mg_d - \frac{m_d v^2}{r} \] where \( m_d = 72 \) kg is the driver's mass. Substitute the known values:\[ N_d = (72 \times 9.81) - \frac{72 \times (12)^2}{88} \] Calculate:\[ N_d = 706.32 - 118.36 = 587.96 \text{ N} \] Thus, the normal force exerted by the car on the driver is 587.96 N.

Step 5: Solve for Part (c) - Speed for Zero Normal Force

To find the car speed where the normal force on the driver equals zero, set the normal force equation to zero:\[ 0 = mg_d - \frac{m_d v^2}{r} \] Solve for \( v \):\[ mg_d = \frac{m_d v^2}{r} \] \[ v^2 = rg_d \] \[ v = \sqrt{rg_d} \] Substitute the known values:\[ v = \sqrt{88 \times 9.81} \] Calculate:\[ v = \sqrt{863.28} \approx 29.38 \text{ m/s} \] Therefore, the car speed at which the normal force on the driver becomes zero is approximately 29.38 m/s.

Key Concepts

Normal Force

The normal force is a contact force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In this scenario, as the sports car crosses the hilltop, the normal force acts upward against the force of gravity pulling the car down. At the top of the hill, the car’s weight tries to push it downwards due to gravity, while the normal force decreases because the car is in a curved path.

• When an object moves in a circular motion on top of a hill, the normal force is calculated using the equation: \[N = mg - \frac{mv^2}{r}\]This equation states that the normal force, \(N\), is the difference between the gravitational force \(mg\), and the centripetal force \(\frac{mv^2}{r}\) required to keep the object moving in its circular path.
• In the given exercise, the calculated normal force on the car when at the top of the hill is 7973.84 N, meaning the road exerts this force upward on the car.
• For the driver, the normal force is calculated similarly, leading to a force exerted on him of 587.96 N.

Normal force changes with speed; faster speeds reduce it because the centrifugal effect increases, requiring less support from the surface.

Gravitational Force

Gravitational force is that force which attracts two bodies towards each other. It is the force the Earth exerts on objects pulling them downward, which is given by the formula \(F = mg\). Here, \(m\) is the mass and \(g\) is the acceleration due to gravity, approximately \(9.81\,\text{m/s}^2\).

• The force is directed towards the center of the Earth, providing the necessary centripetal force required for circular motion as the car goes over the hill.
• In this exercise, the gravitational force on the car is calculated as \(mg = 975 \times 9.81\), which results in 9564.75 N pressing downwards.
• Similarly, the force on the driver is determined:\(mg_d = 72 \times 9.81\), giving a downward force of 706.32 N.

Gravitational force remains constant unless the mass or location changes, unlike normal force which can vary with speed and motion.

Circular Motion

Circular motion occurs when an object moves along a path in the shape of a circle. In this exercise, circular motion is highlighted when the sports car travels over the top of a hill. This motion is maintained by centripetal forces that direct towards the center of the circular path.

• The force required for circular motion is provided by components of weight when at the top of the hill, since there is no additional surface or applied force helping in lateral direction.
• Centripetal force is given by \(F_c = \frac{mv^2}{r}\), ensuring the car stays in its circular path without flying off. This is balanced by the combination of gravitational and normal forces.
• When the speed of the car increases to the point where the normal force becomes zero, which happens at the critical velocity \(v\) as calculated: \(v = \sqrt{rg_d} \approx 29.38 \text{ m/s}\).This means at this speed, entire centripetal requirement is fulfilled by gravity with no additional support from the road.

Circular motion is all about balancing forces to maintain the path, showcasing the interplay between different forces.