A father fashions a swing for his children out of a long rope that he fastens to
ID: 1539371 • Letter: A
Question
A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children swings from this rope that is 10 m long, his tangential speed at the bottom of the swing is 8 m/s. (ignore the mass of the rope)
(a) What is the centripetal acceleration of the child at the bottom of the swing?
m/s2
(b) What is the tension in the rope at the bottom if the child's mass is 50 kg?
N
c) If the swing is stopped and is no longer moving what would the tension in the rope be?
N
TPExplanation / Answer
(a)
centripetal acceleration ac = v^2/L
L = length of the rope
v = speed at the bottom
ac = 8^2/10 = 6.4 m/s^2
======================
(b)
along vertical
Fnet = T - mg
from newtons second law Fnet = m*ac
T - mg = m*ac
T = m*g + m*ac
T = (50*9.8) + (50*6.4) = 810 N <<<<====answer
=======================
if the swing stops v = 0
T = mg = 50*9.8 = 490 N <<<<=====answer
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