Question

As your bus rounds a flat curve at constant speed, a package with mass 0.300kg , suspended from the luggage compartment of the bus by a string 50.0cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0? with the vertical. In this position, the package is 40.0mfrom the center of curvature of the curve.

What is the speed of the bus?

Answer 1

Fix frame of reference on the bus and consider the lunch box as the system to be studied.
Mass of the lunch box m = 0.300 kg
Radius of curvature r = 40.0 m
Let the speed of the bus = v

Forces on the lunch box are: -
1. Tension T in the string at angle 30.0 deg with vertical
2. Weight mg downward
3. Centrifugal force mv^2/r radially outward

In the frame of reference of the bus, the lunch box is in equilibrium. Therefore, net force on the lunch box = 0
Taking vertical components of forces,
T cos(30.0 deg) - mg = 0
Or T = mg/cos(30.0 deg)---------------------------(1)

Taking horizontal components of forces,
T sin(30.0 deg) - mv^2/r = 0
Or T = (mv^2/r)/sin(30.0 deg)--------------------(2)

From (1) and (2),
mg/cos(30.0 deg) = (mv^2/r)/sin(30.0 deg)
Dividing both sides by m,
g/cos(30.0 deg) = (v^2/r)/sin(30.0 deg)
Or g * sin(30.0 deg)/cos(30.0 deg) = v^2/r
Or g*tan(30.0 deg) = v^2/r
Or v^2 = rg*tan(30.0 deg)
Or v = sqrt{rg*tan(30.0 deg)}
Or v = sqrt(40*9.81*0.58)
Or v = sqrt(227.592)
Or v = 15.09 m/s