Problem 96 - Physics for Scientists and Engineers with Modern Physics

Question

Consider a train that rounds a curve with a radius of \(570 \mathrm{~m}\) at a speed of \(160 \mathrm{~km} / \mathrm{h}\) (approximately \(100 \mathrm{mi} / \mathrm{h}\) ). ( \(a\) ) Calculate the friction force needed on a train passenger of mass \(75 \mathrm{~kg}\) if the track is not banked and the train does not tilt. (b) Calculate the friction force on the passenger if the train tilts at an angle of \(8.0^{\circ}\) toward the center of the curve.

Step-by-Step Solution

Answer

(a) 259.1 N, (b) 250.5 N

Step 1: Convert Speed to Meters per Second

To calculate the friction force, first convert the speed from kilometers per hour to meters per second. We use the conversion factor: \(1\, \mathrm{km/h} = \frac{1}{3.6}\, \mathrm{m/s}\).So, \(160\, \mathrm{km/h} = \frac{160}{3.6}\, \mathrm{m/s} \approx 44.44\, \mathrm{m/s}\).

Step 2: Calculate Centripetal Force Required

The centripetal force \(F_c\) required for uniform circular motion is given by the formula:\[F_c = \frac{mv^2}{r}\]Substituting the given values, \(m = 75\, \mathrm{kg}\), \(v = 44.44\, \mathrm{m/s}\), and \(r = 570\, \mathrm{m}\):\[F_c = \frac{75 \times (44.44)^2}{570} \approx 259.1\, \mathrm{N}\]

Step 3: Calculate Friction Force for Unbanked Track

For (a), on an unbanked track, the only horizontal force acting is the frictional force, which must provide the entire centripetal force. Thus, the friction force \(f = F_c\):\[f = 259.1\, \mathrm{N}\]

Step 4: Calculate Friction Force for Banked Track

For (b), with the train tilting at an angle \(\theta = 8.0^\circ\), the normal force has a horizontal component contributing to the centripetal force. The friction force \(f\) is adjusted as follows:- The normal force \(N = mg\cos(\theta)\). - Horizontal component of the normal force: \(N\sin(\theta) = mg\cos(\theta)\sin(\theta)\).Thus, the total centripetal force is the sum of this horizontal component and the required friction force:\[f + mg\cos(\theta)\sin(\theta) = \frac{mv^2}{r}\]Solving for \(f\):\[f = \frac{mv^2}{r} - mg\cos(\theta)\sin(\theta)\]Substitute \(m = 75\, \mathrm{kg}\), \(v = 44.44\, \mathrm{m/s}\), \(r = 570\, \mathrm{m}\), \(g = 9.81\, \mathrm{m/s^2}\), and \(\theta = 8^\circ\):\[f = 259.1 - 75 \times 9.81 \times \cos(8^\circ) \times \sin(8^\circ)\]\[f \approx 250.5\, \mathrm{N}\]

Key Concepts

Centripetal Force

Centripetal force is key when discussing circular motion. It's the force that keeps objects moving in a circle, directing them towards the center of the path. Without it, objects would continue straight due to inertia. This force is crucial for scenarios such as a car turning on a curvy path or a satellite orbiting Earth.

Understanding the Formula
- The formula for centripetal force is given by \(F_c = \frac{mv^2}{r}\). This formula tells us the force needed to keep an object moving in a circle.

• \(m\) stands for mass. It's how much stuff is in the object, measured in kilograms.
• \(v\) is velocity or how fast the object is moving in meters per second.
• \(r\) is the radius of the path in meters, or how big the circle is.

Practical Example
In the context of a train on a curved track, the centripetal force is vital for keeping it safely on the track. If the train's speed or the curve's radius changes, the amount of force required for safe travel also changes.

Friction Force

Friction force acts as a braking and holding force between surfaces. It's why we don't slip when we walk and why cars can navigate bends without skidding off the road.

Role in Circular Motion
When a train or vehicle goes around a corner, friction supplies some or all of the centripetal force necessary to keep it on track. On a flat, unbanked surface, like in our exercise, the friction force has to match the centripetal force.

• This means on an unbanked curve, the friction force equals the centripetal force (\(f = F_c\)).
• This requirement puts a limit on how fast the train can go without slipping.

Influence of Surface Conditions
The amount of friction force available depends on the surface condition and the materials in contact. If a train tilts slightly, the situation changes because some other forces also begin to help in providing the required centripetal force.

Circular Motion

Circular motion is when an object moves in a circle or along a curved path. Its distinctive feature is the constant change in direction making it different from straight-line motion.

Forces Involved
- To stay in circular motion, an object must consistently be pushed or pulled towards the center of the circle. This is where centripetal force comes in, acting as the inward force needed.
- Friction can provide this inward force, especially on objects like vehicles turning on unbanked tracks.

Real-Life Application
Imagine riding a train moving through a curve. The train continuously changes direction to follow the track. That change requires forces directed towards the center of the curve, preventing it from veering off.
- The train’s motion is a great example of circular motion, showcasing how these forces function in a practical setting.