A wheel 1.70 m in diameter lies in avertical plane and rotates with a constant a

Question

A wheel 1.70 m in diameter lies in avertical plane and rotates with a constant angular acceleration of4.20 rad/s2. The wheelstarts at rest at t = 0, and the radius vector of acertain point P on the rim makes an angle of 57.3°with the horizontal at this time. At t = 2.00 s find thefollowing. (a) the angular speed of the wheel
1 rad/s

(b) the tangential speed and the total acceleration of the pointP
2 m/s
3 m/s2

(c) the angular position of the point P
4 rad (a) the angular speed of the wheel
1 rad/s

(b) the tangential speed and the total acceleration of the pointP
2 m/s
3 m/s2

(c) the angular position of the point P
4 rad

Explanation / Answer

diameter d= 1.70m radius r= d/ 2 = 0.85 m
angular acceleration = 4.20 rad/s2. Initial angular speed w = 0 Initial angle = 57.3°                      = 57.3 * ( / 180 )                      = 1 rad time t= 2.00 s (a) the angular speed of the wheel after 2s is w ' = w + t                                                                          = 0 + (4.2 * 2 ) = 8.4 rad / s

(b) the tangential speed at time 2 s is v = r w '                                                             =7.14 m / s the total acceleration of the point P a = [ a ' ^ 2 + a" ^ 2 ] where a ' = radial accleration = v ^ 2/r = 59.976 m / s ^ 2           a " =Tangential accleration = r = 3.57 m / s ^ 2 Plug the values we get totalaccleration a = 60.08 m / s ^ 2

(c) the angular position of the point P is ' = + " where " = angulardisplacement in time t = 2 s                = wt + ( 1/ 2) t ^ 2                = 0 + 8.4                = 8.4 rad So, ' = 1 rad + 8.4 rad = 9.4rad

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