Respuesta :
Answer:
Angular acceleration = 6.37rad/sec²
Approximately, Angular acceleration =
6.4 rad/sec²
Explanation:
Length of the rod = 2.0m long
Inclination of the rod (horizontal) = 30°
Mass of the rod is not given so we would refer to it as = M
Rotational Inertia of the Rod(I) = 1/3ML²
Angular Acceleration = ?
There is an equation that shows us the relationship between Torque and Angular acceleration.
The equation is :
Torque(T) = Inertia × Angular Acceleration
Angular acceleration = Torque ÷ Inertia
Where:
Torque = L/2(MgCosθ)
Where M = Mass
L = Length = 2.0m
θ = Inclination of the rod (horizontal) = 30°
g = Acceleration due to gravity = 9.81m/s²
Inertia = 1/3ML²
Angular Acceleration = (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²
Angular Acceleration =
(3 × g × cos 30°) ÷ 2× L
Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L
Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m
Angular Acceleration = 6.37rad/sec²
Approximately Angular Acceleration =
6.4rad/sec²
The angular acceleration of the rod at the instant it is released is 6.37 [tex]rad/s^2[/tex].
Given the following data:
- Length = 2 meter
- Angle = 30°
To calculate the angular acceleration of the rod at the instant it is released:
The moment of inertia of the rod about a horizontal, frictionless pin through one end (axis) is given by the formula:
[tex](\frac{1}{3})ML^2[/tex] ...equation 1.
Mathematically, torque is given by this formula:
[tex]T = I\alpha[/tex] ....equation 2.
Where:
- T represents torque.
- I is the moment of inertia.
- [tex]\alpha[/tex] is the angular acceleration.
Making [tex]\alpha[/tex] the subject of formula, we have:
[tex]\alpha =\frac{T}{I}[/tex] ....equation 3.
For the horizontal component, torque is given by the formula:
[tex]Torque = \frac{L}{2}(MgCos\theta)[/tex] ...equation 4
Substituting eqn 1 and eqn 4 into eqn 3, we have:
[tex]\alpha =\frac{\frac{L}{2}(MgCos\theta)}{(\frac{1}{3})ML^2} \\\\\alpha = \frac{\frac{MgCos\theta L}{2}}{\frac{ML^2}{3}} \\\\\alpha = \frac{MgCos\theta L}{2} \times \frac{3}{ML^2} \\\\\alpha = \frac{3gCos\theta}{2L} \\\\\alpha =\frac{\frac{L}{2}(MgCos\theta)}{(\frac{1}{3})ML^2} \\\\\alpha = \frac{\frac{MgCos\theta L}{2}}{\frac{ML^2}{3}} \\\\\alpha = \frac{MgCos\theta L}{2} \times \frac{3}{ML^2} \\\\\alpha = \frac{3gCos\theta}{2L} \\\\\alpha = \frac{3\;\times \;9.8 \times\; Cos30}{2\times 2}[/tex]
[tex]\alpha = \frac{29.4 \times\; 0.8660}{4}\\\\\alpha = \frac{25.46}{4}[/tex]
Angular acceleration, [tex]\alpha[/tex] = 6.37 [tex]rad/s^2[/tex]
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