Angular position and velocity of a wheel (7424)

, por F_y_Q

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The angular velocity of a bicycle wheel is 5\ rad\cdot s^{-1}, and its angular acceleration is 3\ rad\cdot s^{-2}.

a) What are the angular position and angular velocity at t = 5 s?

b) What are the angular position and angular velocity at t = 5 s, expressed in revolutions?

c) What are the final velocity and displacement of the bicycle at t = 5 s if the tire has a diamater of 1 meter?

P.-S.

a) The equation for angular velocity is:

\color[RGB]{2,112,20}{\bm{\omega= \omega_0 + \alpha\cdot t}}

Using the given values:

\omega = 5\ \frac{rad}{s} + 3\ \frac{rad}{s\cancel{^2}}\cdot 5\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bm{20\ rad\cdot s^{-1}}}}


The equation for angular position is:

\color[RGB]{2,112,20}{\bm{\varphi= \omega_0\cdot t + \frac{\alpha}{2}\cdot t^2}}

Using the given values:

\varphi = 5\ \frac{rad}{\cancel{s}}\cdot 5\ \cancel{s} + \frac{3}{2}\ \frac{rad}{\cancel{s^2}}\cdot 5^2\ \cancel{s^2} = \fbox{\color[RGB]{192,0,0}{\bf 63 \ rad}}


b) To convert angular velocity and position to revolutions:

\omega = 20\ \frac{\cancel{rad}}{s}\cdot \frac{1\ rev}{2\pi\ \cancel{rad}} = \fbox{\color[RGB]{192,0,0}{\bm{3.2\ rev\cdot s^{-1}}}}


For angular position:

\varphi = 63\ \cancel{rad}\cdot \frac{1\ rev}{2\pi\ \cancel{rad}} = \fbox{\color[RGB]{192,0,0}{\bf 10 \ rev}}


c) To convert angular quantities into linear quantities, use the wheel radius:

v = \omega\cdot R = 20\ \frac{1}{s}\cdot 0.5\ m = \fbox{\color[RGB]{192,0,0}{\bm{10\ m\cdot s^{-1}}}}


d = \varphi\cdot R = 63\cdot 0.5\ m= \fbox{\color[RGB]{192,0,0}{\bf 32\ m}}


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