A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
1 answer:
Answer:
v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²
Explanation:
This is a missile throwing exercise.
On the x axis there is no acceleration so the velocity on the x axis is constant
v₀ₓ = 10 m / s
On the y-axis velocity is affected by the acceleration of gravity, let's use the equation
v_y = - g t
at the highest point of the trajectory the vertical speed must be zero
v_y = 0
therefore the velocity of the body is
v = (10 i ^ + 0j ^) m / s
the acceleration is
a = (0 i ^ - g j⁾
a = (0i ^ - 9.8 j ^) m / s²
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Answer:
≈ 2.1 R
Explanation:
The moment of inertia of the bodies can be calculated by the equation
I = ∫ r² dm
For bodies with symmetry this tabulated, the moment of inertia of the center of mass
Sphere = 2/5 M R²
Spherical shell = 2/3 M R²
The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass
I = + M D²
Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation
Let's start with the spherical shell, axis is along a diameter
D = 2R
Ic = + M D²
Ic = 2/3 MR² + M (2R)²
Ic = M R² (2/3 + 4)
Ic = 14/3 M R²
The sphere
Is = + M [
²
Is = Ic
2/5 MR² + M ² = 14/3 MR²
² = R² (14/3 - 2/5)
= √ (R² (64/15)
= 2,066 R