A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
1 answer:
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
You might be interested in
Answer:
E =
Explanation:
For this exercise let's use Gauss's law. The Gaussian surface that follows the symmetry of the charges is a sphere
Ф = ∫ E. dA =
the bold are vectors, the radii of the sphere and the electric field are parallel therefore the scalar product reduces to the algebraic product
Ф = ∫ E dA = \frac{x_{int} }{\epsilon_o}
E ∫ dA = \frac{x_{int} }{\epsilon_o}
E A = \frac{x_{int} }{\epsilon_o}
the area of a sphere is
A = 4π r²
the charge inside the sphere is q = + q
we substitute
E 4π r² = \frac{x }{\epsilon_o}
E =