A pendulum with a cord of length r = 0.950 m swings in a vertical plane (see fig

Question

A pendulum with a cord of length

r = 0.950 m

swings in a vertical plane (see figure). When the pendulum is in the two horizontal positions

= 90°

and

= 270°,

its speed is 5.00 m/s.

(a) Find the magnitude of the radial acceleration for these positions.

________m/s2

(b) Find the magnitude of the tangential acceleration for these positions.
________m/s2

(c) Calculate the magnitude and direction of the total acceleration.

Explanation / Answer

a)
ar = v^2/r
= 5^2/0.95
=26.3 m/s^2
For both positions

b)
at = g*sin
at = 90°
at = g*sin
=g*sin 90
= g
= 9.8 m/s^2

at = 270°
at = g*sin
=g*sin 270
= -g
= -9.8 m/s^2

c)
at = 90°
magnitude = sqrt (26.3^2 + 9.8^2) = 28.1 m/s^2
direction = atan (at/ar) = atan (9.8/26.3) = 20.4 degree

at = 270°
magnitude = sqrt (26.3^2 + 9.8^2) = 28.1 m/s^2
direction = atan (at/ar) = atan (-9.8/26.3) = -20.4 degree

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