(637) A cyclist is riding on a horizontal frozen meadow. He is turning so that he is tilted at an angle of 60°. Radius of curvature of the bend is 100m. What is the maximum speed of the cyclist?
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Since you do not specify, I will assume 60 degrees from horizontal
tan 60 = mg/(m v^2/r)
v^2/r = g/tan 60
v^2 = g r/tan 60
v = sqrt(981/tan 60)
v = 23.8 m/s
To find the maximum speed of the cyclist, we need to consider the forces acting on the cyclist during the turn.
In this case, the cyclist is subject to two main forces: the force of gravity (pulling the cyclist downwards) and the force of friction between the bicycle tires and the icy surface (providing the necessary centripetal force for the cyclist to make the turn).
To determine the maximum speed at the given radius of curvature (100m) and angle of tilt (60°), we can use the following steps:
Step 1: Resolve forces vertically and horizontally.
- The weight of the cyclist can be resolved into two components: one perpendicular to the surface (N), and one parallel to the surface (mg sin 60°).
- The friction force (f) acts horizontally and inward, providing the centripetal force required to keep the cyclist on a curved path.
Step 2: Equate the friction force and the centripetal force.
- The friction force (f) provides the centripetal force needed for the cyclist to make the turn.
- The centripetal force (Fc) can be calculated using the formula: Fc = m