Answer to Question #102740 in Mechanics | Relativity for Ahmad

Question #102740
A wheel 2.0 m in diameter lies in the vertical plane and rotates about its central axis
with a constant angular acceleration of 4.0rads .
2
The wheel starts from rest at
t  0 and the radius vector of a point A on the wheel makes an angle of 60º with the
horizontal at this instant. Calculate the angular speed of the wheel, the angular
position of the point A and the total acceleration at t  2.0s.
1
Expert's answer
2020-02-12T10:00:48-0500

1) The angular speed of the wheel


"\\omega=\\omega_0+\\epsilon t=0+\\epsilon t=\\epsilon t=4\\cdot 2=8 rad\/s"


2) The angular position of the point A


"\\phi=\\phi_0+\\omega_0+\\frac{\\epsilon t^2}{2}=\\frac{\\pi}{3}+0+\\frac{4\\cdot 2^2}{2}=1.047+8=9.047 rad"


or


"\\phi=518.4\u00b0"



"518.4\u00b0-360\u00b0=158.4\u00b0"with the horizontal


3) The total acceleration


"a=\\sqrt{a^2_n+a^2_\\tau}=\\sqrt{(\\frac{v^2}{R})^2+\\epsilon^2R^2}=\\sqrt{(\\frac{\\omega^2\\cdot R^2}{R})^2+\\epsilon^2R^2}=\\sqrt{(\\omega^2\\cdot R)^2+\\epsilon^2R^2}==\\sqrt{(8^2\\cdot 1)^2+4^2\\cdot 1^2}=64.1m\/s^2"










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