Explanation:
It is given that,
Mass of the brick, m = 1.15 kg
Radius of the circle, r = 1.44 m
The cable will break if the tension exceeds 43.0 N
Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,




v = 6.30 m/s
So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.
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Answer:
J = Δp
Explanation:
The impulse-momentum theorem says that the impulse J is equal to the change in momentum p.
J = Δp
Answer:
The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.
Explanation:
Maximum speed is :
v (max) = Aω
Speed v at any displacement y is given by
=
(
-
) ........................................................ i
And,
v =
v (max)
or, 2 × v = Aω .................................................... ii
Eliminating ω from equations i and ii,
=
(
-
)
or,
= (
)
=(
) 
or, y = 3.56 cm.