Answer:
α = (ω²)/8π
Explanation:
The angular acceleration(α) of the carousel can be determined by using rotational
kinematics:
ω² =ωo² + 2αθ
Let's make α the subject of this equation ;
ω² - ωo² = 2αθ
α = (ω² −ωo²)/2θ
Now, from the question, since initially at rest, thus, ωo = 0
Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π
Plugging in the relevant values to get ;
α = (ω²)/2(4π)
α = (ω²)/8π
The correct answer is 17.24 m/s. You get the answer by subtracting the two heights of the tracks which are 36.5 and 10.8 m, and the answer is 25.7. Since you already know the height at which the kinetic energy will be coming from, you then divide the amount of weight the roller coaster has to the distance it needs to travel in order for you to determine the speed of the car. So that is, 4,357 kg and 25.7 m and the answer is 169 kg/m. Dividing it to the earth's gravity of 9.8 m/s you'll get 17.24 m/s.
<span>As it is descended from a vertical height h,
The lost Potential Energy = Mgh
The gained Kenetic Energy = (1/2)Mv^2; The rotational KE = (1/2)Jw^2
The angular speed w = speed/ Radius = v/R
So Rotational KE = (1/2)Jw^2 = (1/2)J(v/R)^2; J is moment of inertia
Now Mgh = (1/2)Mv^2 + (1/2)J(v/R)^2 => 2gh/v^2 = 1 + (J/MR^2)
As v = (5gh/4)^1/2, (J/MR^2) = 2gh/v^2 - 1 => (J/MR^2) = (8gh/5gh) - 1
so (J/MR^2) = 3/5 and therefore J = (3/5)MR^2.</span>
Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
