The rigid object shown in Fig. 10-63 consists of three balls and three connecting rods, with M = 1.5 kg, L = 0.67 m, and θ = 32°. The balls may be treated as particles, and the connecting rods have negligible mass. Determine the rotational kinetic energy of the object if it has an angular speed of 1.2 rad/s about
(a) an axis that passes through point P and is perpendicular to the plane of the figure, and
(b) an axis that passes through point P, is perpendicular to the rod of length 2L, and lies in the plane of the figure.
The figure shows a point P, then to the left of that point is a ball of mass 2M connected by a rod length L at an angle theta degrees from the horizontal. There is another mass 2M ball that is connected by a rod of length L at negative theta to the horizontal. To the right of point P there is a ball of mass M connected by a rod of length 2L.
So far, I have the formula K-.5Iw2, but I don't know how to find I for this system of particles. Any help would be much appreciated
The system moment of inertia will be the sum of all three. If the masses are the same, and the lengths the same, 3/2mr^2 will be total. Remember, when you derived the moment of inertia, you integrated (summed) all point masses at some distance r from the rotation.
Okay so I got the first part alright, but then when the axis of rotation changes, would that affect I?
To find the rotational kinetic energy of the object, we need to determine the moment of inertia of the system. The moment of inertia depends on the distribution of mass and the axis of rotation.
(a) To find the moment of inertia for rotation about an axis passing through point P and perpendicular to the plane of the figure, we can use the parallel axis theorem. This theorem states that the moment of inertia for an object rotating about an axis parallel to and a distance "d" away from an axis through its center of mass is given by:
I = Icm + Md^2
Where Icm is the moment of inertia about an axis passing through the center of mass and Md^2 represents the contribution of the object's mass to the moment of inertia due to its distance from the axis of rotation.
For our object, we can treat the balls as particles, so their moment of inertia about an axis passing through their center of mass is given by:
Icm = m*r^2
Where m is the mass of the ball and r is its distance from the axis of rotation.
Let's calculate the moment of inertia for each ball:
For the ball of mass 2M on the left:
I1 = (2M)*((0.67*cos(θ/2))^2)
For the ball of mass 2M on the right:
I2 = (2M)*(2*0.67*cos(θ))^2
For the ball of mass M in the middle:
I3 = M*((2*0.67*sin(θ))^2)
Now, we can use the parallel-axis theorem to find the total moment of inertia of the system:
I_total = I1 + I2 + I3
Once you have the moment of inertia (I_total), you can use the formula for rotational kinetic energy:
K = 0.5 * I_total * ω^2
Where ω is the angular speed.
(b) To find the moment of inertia for rotation about an axis passing through point P, perpendicular to the rod of length 2L, and lying in the plane of the figure, we need to consider the moment of inertia of each component about this axis individually.
For the ball of mass 2M on the left, the distance between the axis of rotation and the ball is L. So its moment of inertia about this axis is:
I1 = (2M)*(L^2)
For the ball of mass 2M on the right, the distance between the axis of rotation and the ball is 2L. So its moment of inertia about this axis is:
I2 = (2M)*((2L)^2)
For the ball of mass M in the middle, the distance between the axis of rotation and the ball can be found using trigonometry. Let's call it d. We can express d in terms of the angle θ and the length of the rod 2L:
d = 2L*sin(θ/2)
So its moment of inertia about this axis is:
I3 = M*(d^2)
Again, we can calculate the total moment of inertia using the parallel-axis theorem:
I_total = I1 + I2 + I3
And then use the rotational kinetic energy formula:
K = 0.5 * I_total * ω^2
I hope this helps! Let me know if you have any further questions.