Calculate the gravitational force between a 10 kg ball and 20 kg ball
1 answer:
Answer:
Explanation:
The gravitational force is an attractive force exerted between any objects with mass.
The magnitude of the gravitational force between two objects is given by:
where
G is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between the two objects
In this problem, we have:
is the mass of the 1st ball
is the mass of the 2nd ball
r = 5 m is the separation
So, the gravitational force between them is
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The radius of the Ferris wheel is r = 83/2 = 91.5 m.
The angular velocity is
ω = (2π rad)/(37.3*60 s) = 2.8075 rad/s
The tangential velocity is
v = rω = 0.2569 m/s
The arc length traveled in 8.6 min (= 8.6*60 s = 516 s) is
s = (91.5 m)*(2.8075 rad/s)*(516 s) = 132.553 m
The central angle swept is
θ = 132.553/91.5 = 1.4487 rad
From the vector diagram, the change in velocity is (from the Law of Cosines)
Δv² = v²(1 - 2 cosФ)
where Ф = Π - 1.4487 = 1.6929 rad
Δv² = 0.2569²[1 - 2*(-0.1218)] = 0.0821
Δv = 0.2865 m/s
The acceleration is
a₁ = (0.2865 m/s)/(516 s) = 5.6 x 10⁻⁴ m/s²
The actual centripetal acceleration is directed toward the center of the wheel, and its value is
a = v²/r = 0.2569²/91.5 = 7.2 x 10⁻⁴ m/s²
Answer:
a = 7.2 x 10⁻⁴ m/s², the centripetal acceleration acting toward the center of the wheel.
The magnitude of a₁ is 5.6 x 10⁻⁴ m/s², but it is not directed toward the center of the wheel.
The angular velocity is
ω = (2π rad)/(37.3*60 s) = 2.8075 rad/s
The tangential velocity is
v = rω = 0.2569 m/s
The arc length traveled in 8.6 min (= 8.6*60 s = 516 s) is
s = (91.5 m)*(2.8075 rad/s)*(516 s) = 132.553 m
The central angle swept is
θ = 132.553/91.5 = 1.4487 rad
From the vector diagram, the change in velocity is (from the Law of Cosines)
Δv² = v²(1 - 2 cosФ)
where Ф = Π - 1.4487 = 1.6929 rad
Δv² = 0.2569²[1 - 2*(-0.1218)] = 0.0821
Δv = 0.2865 m/s
The acceleration is
a₁ = (0.2865 m/s)/(516 s) = 5.6 x 10⁻⁴ m/s²
The actual centripetal acceleration is directed toward the center of the wheel, and its value is
a = v²/r = 0.2569²/91.5 = 7.2 x 10⁻⁴ m/s²
Answer:
a = 7.2 x 10⁻⁴ m/s², the centripetal acceleration acting toward the center of the wheel.
The magnitude of a₁ is 5.6 x 10⁻⁴ m/s², but it is not directed toward the center of the wheel.