As your bus rounds a flat curve at constant speed, a package with mass .5 kg, suspended from the luggage compartment of the bus by a string 45 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30 degrees with the vertical. In this position, the package is 50 m from the center of curvature of the curve. What is the speed of the bus?
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To find the speed of the bus, we can start by analyzing the forces acting on the package.
1. Draw a Free Body Diagram:
- The gravitational force acting on the package is straight down (vertical).
- The tension force in the string is along the string, making an angle of 30 degrees with the vertical.
- There is also the centripetal force acting towards the center of curvature of the curve.
2. Analyze the forces:
- The gravitational force (mg) can be resolved into two components: mg*sin(30) acting along the string and mg*cos(30) acting vertically downward.
- The tension force can be resolved into two components: T*sin(30) acting vertically upward and T*cos(30) acting towards the center of curvature.
3. Equate the forces:
- Equate the vertical components of the forces: T*sin(30) = mg*cos(30).
- Equate the horizontal components of the forces: T*cos(30) = (mv^2) / r, where v is the speed of the bus and r is the radius of curvature.
4. Solve the equations:
- From the vertical component equation, we get T = (mg*cos(30)) / sin(30) = mg.
- Substituting T in the horizontal component equation, we get mg*cos(30) = (mv^2) / r.
- Simplifying, v^2 = (g * r * cos(30)) / sin(30).
- Since g is the acceleration due to gravity (approximately 9.8 m/s^2) and r is given as 50 m, we can substitute these values into the equation.
- v^2 = (9.8 * 50 * cos(30)) / sin(30).
5. Calculate the speed:
- Substituting the values and simplifying, we get v^2 = 245.
- Taking the square root of both sides, we find that v ≈ 15.65 m/s.
Therefore, the speed of the bus is approximately 15.65 m/s.