Certainly! Let's break down the problem into detailed steps addressing each part systematically.
### [C2] a) Drawing a Free Body Diagram (FBD)
Since we cannot draw diagrams here, let's describe the elements and their direction:
- Identify Forces:
1. Normal Force (N): Acts perpendicular to the surface of the banked curve.
2. Gravitational Force (mg): Acts vertically downward.
3. There is no frictional force involved because we are finding the ideal speed in the absence of friction.
- Axis Setup:
- Tilt the x-axis along the incline of the banked curve, such that x-axis is parallel to the plane and y-axis is perpendicular to it.
- Let θ be the banking angle of 5.7°.
### [A3] b) Resolve the forces into x- and y-components
1. Gravitational Force (mg):
- In the direction perpendicular to the bank: [tex]\( mg \cos(\theta) \)[/tex]
- Parallel to the bank: [tex]\( mg \sin(\theta) \)[/tex]
2. Normal Force (N):
- Parallel to the surface: No component because it acts perpendicular to the surface.
- Perpendicular to the surface: [tex]\(N\)[/tex]
### [T4] c) Find the speed.
To find the ideal speed [tex]\(v\)[/tex] of the train rounding the curve, we need to set up the conditions for circular motion on a banked curve without friction.
- Since it is an ideal case (no friction), the centripetal force required for circular motion is provided by the horizontal component of the normal force.
The centripetal force [tex]\(F_c\)[/tex] required for circular motion is given by:
[tex]\[ F_c = \frac{mv^2}{r} \][/tex]
where:
[tex]\( r \)[/tex] is the radius of the curve,
[tex]\( m \)[/tex] is the mass of the train, and
[tex]\( v \)[/tex] is the velocity to be determined.
The component of the normal force that provides this centripetal force is:
[tex]\[ N \sin(\theta) \][/tex]
The vertical forces must balance for the train not to move vertically. This balance gives:
[tex]\[ N \cos(\theta) = mg \][/tex]
[tex]\[ N = \frac{mg}{\cos(\theta)} \][/tex]
Substituting this in the horizontal force equation:
[tex]\[ N \sin(\theta) = \frac{mv^2}{r} \][/tex]
[tex]\[ \frac{mg \sin(\theta)}{\cos(\theta)} = \frac{mv^2}{r} \][/tex]
Notice that the mass [tex]\(m\)[/tex] cancels out:
[tex]\[ g \frac{\sin(\theta)}{\cos(\theta)} = \frac{v^2}{r} \][/tex]
[tex]\[ g \tan(\theta) = \frac{v^2}{r} \][/tex]
Solving for [tex]\(v\)[/tex]:
[tex]\[ v = \sqrt{r \cdot g \cdot \tan(\theta)} \][/tex]
Plugging in the given values:
- [tex]\( r = 5.5 \times 10^2 \, \text{m} \)[/tex]
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- [tex]\( \theta = 5.7^\circ \)[/tex]
First, convert the angle to radians:
[tex]\[ \theta = 5.7 \times \frac{\pi}{180} = 0.0994 \, \text{radians} \][/tex]
Now compute [tex]\( \tan(5.7^\circ) \)[/tex]:
[tex]\[ \tan(0.0994) = 0.0997 \][/tex]
So,
[tex]\[ v = \sqrt{(5.5 \times 10^2) \cdot 9.8 \cdot 0.0997} \][/tex]
[tex]\[ v = \sqrt{5399.3} \][/tex]
[tex]\[ v \approx 73.48 \, \text{m/s} \][/tex]
### Final Answer:
The ideal speed for the train rounding the curve is approximately [tex]\( 73.48 \, \text{m/s} \)[/tex].
