Angular position and velocity of a wheel (7424)

, por F_y_Q

The angular velocity of a wheel of a bike is 5\ \textstyle{rad\over s}, and its angular acceleration is 3 \ \textstyle{rad\over s^2}.

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

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

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


SOLUCIÓN:

a) The equation of the angular velocity is:

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

Considering the values of the magnitudes:

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


The angular position can be calculated using the equation:

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

Just use problem data:

\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) Using conversion factors in each case:

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


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


c) You have to convert angular quantities into linear quantities and you can do this by using the value of the wheel radius:

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


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


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