Which vector points in the direction of the centripetal acceleration of the plane
1 answer:
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Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 =
cos 30 =
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m (2)
a) we substitute in 1
T =
T =
T = 2.26 N
b) from equation 2
v² =
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² =
v² =
For the problem let us take L = 1 m
let's calculate
v =
v = 1.68 m / s