Because the net radial force is acting as the centripetal force, we can equate them and solve for v: Fradial, net = Fc mv? T-mg cos(45 )= v = (T-mg cos(45')) -T-rg cos(45')

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ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
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Chapter4: Motion In Two And Three Dimensions
Section: Chapter Questions
Problem 64P: A runner taking part in the 200-m dash must run around the end of a track that has a circular arc...
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Why does the m change to an r in the rooted formulas. I want to understand why please.
回 X
a762645ccccS:centripetal-acceleration-and-centrip... & *
First attempt
Taia, net
Because the net radial force is acting as the centripetal force, we can equate
them and solve for v:
Fradial, net = F.
mv²
T-mg cos(45)
v =
(T-mg cos(45'))
m
-T - rg cos(45)
NA
2.5 m
V 3=
1.5 kg
(32 N) - (2.5 m) (9.8 ) cos(45)
m
U 6.001
6.00
4/4
The speed of the mass is 6.00
5:47 PM
63°F
E O 4) ENG
1/16/2022
Transcribed Image Text:回 X a762645ccccS:centripetal-acceleration-and-centrip... & * First attempt Taia, net Because the net radial force is acting as the centripetal force, we can equate them and solve for v: Fradial, net = F. mv² T-mg cos(45) v = (T-mg cos(45')) m -T - rg cos(45) NA 2.5 m V 3= 1.5 kg (32 N) - (2.5 m) (9.8 ) cos(45) m U 6.001 6.00 4/4 The speed of the mass is 6.00 5:47 PM 63°F E O 4) ENG 1/16/2022
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