A pendulum with a cord of length r = 0.850 m swings in a vertical plane (see fig

Question

A pendulum with a cord of length r = 0.850 m swings in a vertical plane (see figure). when the pendulum is in the two horizontal positions theta = 90 degree and theta = 270 degree, its speed 4.40 m/s. (a) Find the magnitude of the radial acceleration for these positions. m/s^2 (b) Find the magnitude of the tangential acceleration for these positions. m/s^2 (c) Draw a vector diagram to determine the direction of the total acceleration for these two positions. (Do this on paper. Your instructor may ask you to turn in this work.) (d) Calculate the magnitude and direction of the total acceleration. magnitude m/s^2 direction degree the horizontal

Explanation / Answer

a) radial acceleration a=v^2/r
so a = 4.40^2/0.850=22.77(m/s2)
b) at these position, the acceleration of the pendulum is a=mg/m=g.
c) downward : g
away from the center : 22.77
d) magnitude
S=sqrt(9.8^2+22.77^2)=24.78(m/s2).
direction with vertical.
tan alpha = 22.77/9.8
so alpha = 66.68(deg)

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