The three 500 g masses in the daigram are connected by massless, rigid rods to form a triangle. What is the triangle's rotational energy (in J) if it rotates at 1.50 rev/s about an axis through the center?
(Diagram shows equilateral triangle with 500g masses at each point, the distance from point to point is 40cm and it has three 60deg angles [equilateral])
To find the rotational energy of the triangle, we need to use the formula for rotational kinetic energy:
Rotational Kinetic Energy = 1/2 * moment of inertia * angular velocity^2
First, let's calculate the moment of inertia of the triangle. The moment of inertia depends on the distribution of mass around the axis of rotation. In this case, the triangle is rotating about an axis passing through its center.
For an equilateral triangle, the moment of inertia can be calculated using the formula:
I = (1/6) * m * a^2
Where:
I is the moment of inertia
m is the mass of each point mass (500g = 0.5kg in this case)
a is the distance from the axis of rotation to any of the triangle's corners (in this case, a = 40cm = 0.4m)
Plugging in the values:
I = (1/6) * 0.5kg * (0.4m)^2
I = 0.0133 kgm^2 (rounded to four decimal places)
Next, we need to calculate the angular velocity in rad/s. Since the question mentions that the triangle rotates at 1.50 rev/s, we can convert revolutions to radians:
1 revolution = 2π radians
So, the angular velocity is:
Angular velocity = 1.50 rev/s * 2π rad/rev
Angular velocity = 9.42 rad/s (rounded to two decimal places)
Now, we can calculate the rotational energy:
Rotational Energy = 1/2 * moment of inertia * angular velocity^2
Rotational Energy = 1/2 * 0.0133 kgm^2 * (9.42 rad/s)^2
Calculating the value:
Rotational Energy ≈ 0.724 J (rounded to three decimal places)
Therefore, the rotational energy of the triangle is approximately 0.724 J.