a point on a rim of a .030 m radius rotating wheel has a tangential speed of 4.0

Question

a point on a rim of a .030 m radius rotating wheel has a tangential speed of 4.0 m/s. What is the tangential speed of a point 0.20 m from the center of the same wheel?

Explanation / Answer

1) if a wheel (radius = r) is rolling on a level surface angular speed (w) then any point on the top of rim v(top of rim) = 2v [v(cm)+ v(rot), both same, parallel] v(bottom of rim) = 0 [v(cm) - v(rot), both same, antiparallel] v(centre) = v [v(cm)] where v = r w ----------------------------------- given V(top, tengential) = v (cm) = r w = 4 m/s w = 4/r = 4/0.3 = 40/3 rad/sec >>>>>> constant rotational speed ---------------------- v( r = 0.20m) = 0.20 * w = 0.20 * 40/3 = 2.67 m/s --------------------------------------… Force of grav = required centripetal force to be in circle of radius r GMm/r^2 = mv^2/r v^2 = GM/r r = R + h = R + R = 2R v^2 = GM/2R v^2 = [GM/R^2] [R/2] = 9.8* R/2 = 4.9 R we know g (earth) = 9.8 = GM/R^2 v^2 = 4.9 * 6.4* 10^6 = 49*64*10^4 v = 7 * 8 * 100 v = 5600 m/s

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