Types of Acceleration in Rotation Video Lessons

Concept

# Problem: A cyclist competes in a one-lap race around a flat, circular course of radius 140m. Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60s. The mass of the bicycle (including the rider) is 76kg. What is the magnitude of the net force acting on the bicycle as it crosses the finish line?

###### FREE Expert Solution

Newton's second law:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$

To get the acceleration a, use the kinematic equation:

$\overline{){\mathbit{s}}{\mathbf{=}}{{\mathbf{v}}}_{{\mathbf{0}}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{a}}{{\mathbf{t}}}^{{\mathbf{2}}}}$

Distance, s = 2πR

$\begin{array}{rcl}\mathbf{2}\mathbf{\pi }\mathbf{R}& \mathbf{=}& {{\mathbf{v}}}_{{\mathbf{0}}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{at}}}^{{\mathbf{2}}}\\ {\mathbf{a}}& \mathbf{=}& \frac{\mathbf{2}\mathbf{\left(}\mathbf{2}\mathbf{\pi R}\mathbf{-}{\mathbf{v}}_{\mathbf{0}}\mathbf{t}\mathbf{\right)}}{{\mathbf{t}}^{\mathbf{2}}}\\ & \mathbf{=}& \frac{\mathbf{2}\mathbf{\left[}\mathbf{\left(}\mathbf{2}\mathbf{\pi }\mathbf{\right)}\mathbf{\left(}\mathbf{140}\mathbf{\right)}\mathbf{-}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{\left(}\mathbf{60}\mathbf{\right)}\mathbf{\right]}}{{\mathbf{60}}^{\mathbf{2}}}\end{array}$

a = 0.489 m/s2

F = (76)(0.489) = 37.2 N

The final speed of the cyclist:

v = v0 + at = 0 + (0.489)(60) = 29.3 m/s

Centripetal force:

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###### Problem Details

A cyclist competes in a one-lap race around a flat, circular course of radius 140m. Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60s. The mass of the bicycle (including the rider) is 76kg.

What is the magnitude of the net force acting on the bicycle as it crosses the finish line?