Jeff of the Jungle swings on a vine that is 7.20 long (see the figure). At the b

Question

Jeff of the Jungle swings on a vine that is 7.20 long (see the figure). At the bottom of the swing, just before hitting the tree, Jeff's linear speed is 8.50 . Suppose that at some point in his swing Jeff of the Jungle has an angular speed of 0.770 and an angular acceleration of 0.650 .(Figure 1)

Part A 1.Find the magnitude of his centripetal acceleration.

2.Find the magnitude of his tangential acceleration.

3.Find the magnitude of his total acceleration.

4.Find the angle his total acceleration makes with respect to the tangential direction of motion.

Jeff of the Jungle swings on a vine that is 7.20 long (see the figure). At the bottom of the swing, just before hitting the tree, Jeff's linear speed is 8.50 . Suppose that at some point in his swing Jeff of the Jungle has an angular speed of 0.770 and an angular acceleration of 0.650 .(Figure 1) Part A 1.Find the magnitude of his centripetal acceleration. 2.Find the magnitude of his tangential acceleration. 3.Find the magnitude of his total acceleration. 4.Find the angle his total acceleration makes with respect to the tangential direction of motion.

Explanation / Answer

1. acp =r*w^ 2
= 7.20 m* 0.770^2 rad/s
= 4.2688 m/s^2

2. at = r*alpha
= 7.20 m * 0.650 rad/s ^2
=4.68 m/s^2

3. a = sqrt(acp^2 + at^2)
= sqrt( 4.2688^2 + 4.68 ^2)
= 6.334 m/s^2

4. Phi = tan-1(acp/ at)
= tan-1(4.2688 /4.68 )
= 42.36 degrees

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