A 0.385-kg rock is swung in a vertical circular path on a string that is 0.480m
ID: 1358704 • Letter: A
Question
A 0.385-kg rock is swung in a vertical circular path on a
string that is 0.480m 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.
What is,
(a) the magnitude aC of the centripetal acceleration at point Q?
(b) the direction of the centripetal acceleration at point Q? Draw an arrow whose tip points in the correct direction.
(c) What is the tension magnitude Tbot in the string at the bottom?
(d) What is the tension magnitude Ttop at the top?
(e) What is the tension magnitude TQ at point Q where the rock is in a horizontal position relative to the center?
Explanation / Answer
a) aC =v^2 / L = 3.90^2 / 0.480 = 31.68 m/s^2
b) Direction of cntripetal acc is alway towards the centre of circle
at point Q
direction -> towards left
c) at the bottom,
T - mg = maC
T = 0.385 ( 9.80 + 31.68) = 15.97 N
d) at top,
T + mg = maC
T = 0.385 ( 31.68 - 9.80) =8.42 N
e) T = maC = 0.385 x 31.68 =12.20 N
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