A 0.385-kg rock is swung in a vertical circular path on a string that is 0.480 m

Question

A 0.385-kg rock is swung in a vertical circular path on a string that is 0.480 m long. Assume the constant speed v of the rock is 3.90 m/s. Three points on the vertical circular path are shown: the top, the bottom and point Q at the end of a horizontal, radial line segment. Assume g = 9.80 m/s2

(a) (15 points) the magnitude aC of the centripetal acceleration at point Q

(b) (5 points) the direction of the centripetal acceleration at point Q? Draw an arrow whose tip points in the correct direction.

(c) (20 points) What is the tension magnitude Tbot in the string at the bottom?

(d) (5 points) What is the tension magnitude Ttop

(e) (5 points) What is the tension magnitude TQ at point Q where the rock is in a horizontal position relative to the center?

Explanation / Answer

a) Cetripetal acceleration = v2/r = (3.9)2 /0.48 = 31.69 m/s/s

b) direction of centripetal acceleration is towards the center

c)when the rock is at bottom Tbot-mg = mv2/r

                                         Tbot = mg +mv2/r = 0.385x9.8 + 0.385x(3.9)2/0.48 = 15.97 N

d) when the rock is at the top Ttop +mg = mv2 /r

                                                 Ttop = mv2/r - mg = 0.385x(3.9)2/0.48 - 0.385x9.8 = 8.43 N

e) when the rock is at Q tension and the weight are perpendicular and weight coponent to Tension =0

                                               Tq = mv2/r = 0.385x(3.9)2/0.48 = 12.2 N

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