A long thin rod is cut into two pieces, one being as long as the other. To the midpoint of piece A, piece B is attached perpendicularly, in order to form the inverted "T" shown in the figure. The application of a net external torque causes this object to rotate about axis 1 with an angular acceleration of 3.6 rad/s2. When the same net external torque is used to cause the object to rotate about axis 2, what is the angular acceleration?
To find the angular acceleration when the object rotates about axis 2, we need to use the principle of conservation of angular momentum. The angular momentum of an object is defined as the product of its moment of inertia and its angular velocity.
In this case, we have an object formed by two pieces. Let's call the moment of inertia of piece A as I_A and the moment of inertia of piece B as I_B. Since both pieces have the same length and are symmetrical, their moment of inertia can be considered the same, so we can represent it as I.
When the object rotates about axis 1, the moment of inertia is given by I_total = I_A + I_B. Since piece B is attached perpendicularly to the midpoint of piece A, the overall shape of the object can be approximated as a straight rod with the length of 2A (the length of piece A) and a perpendicular rod (piece B) attached to it.
The angular acceleration when rotating about axis 1 is given as 3.6 rad/s^2. To find the angular acceleration when rotating about axis 2, we need to calculate the moment of inertia of the object about axis 2.
The moment of inertia about axis 2 can be calculated using the parallel axis theorem, which states that the moment of inertia of an object about any axis parallel to an axis through its center of mass is equal to the moment of inertia about the center of mass plus the mass of the object times the square of the perpendicular distance between the two axes.
In this case, the moment of inertia about axis 2 can be calculated as follows:
I_total2 = I_A + I_B + mB * (2A)^2.
Where mB is the mass of piece B.
Now, since we know that I_total = I_A + I_B and I_total2 = I_A + I_B + mB * (2A)^2, we can rewrite the equation for I_total2 as:
I_total2 = I_total + mB * (2A)^2.
To find the angular acceleration when rotating about axis 2, we can use the principle of conservation of angular momentum, which states that the angular momentum of an object remains constant unless acted upon by an external torque. Since the net external torque acting on the object is the same in both cases, the angular momentum must be conserved.
The angular momentum about axis 1 is given by L1 = I_total * ω1, where ω1 is the angular velocity when rotating about axis 1.
Similarly, the angular momentum about axis 2 is given by L2 = I_total2 * ω2, where ω2 is the angular velocity when rotating about axis 2.
Since the angular momentum is conserved, we can equate L1 and L2:
I_total * ω1 = I_total2 * ω2.
Substituting the expressions for I_total and I_total2, we get:
(I_A + I_B) * ω1 = (I_A + I_B + mB * (2A)^2) * ω2.
Simplifying the equation, we have:
(I_A + I_B) * ω1 = (I_A + I_B) * ω2 + mB * (2A)^2 * ω2.
Dividing both sides by (I_A + I_B), we get:
ω1 = ω2 + [mB * (2A)^2 * ω2] / (I_A + I_B).
Simplifying further, we have:
ω1 - ω2 = [mB * (2A)^2 * ω2] / (I_A + I_B).
Now, we can substitute the given angular acceleration when rotating about axis 1 (3.6 rad/s^2) into the equation:
3.6 rad/s^2 = [mB * (2A)^2 * ω2] / (I_A + I_B).
Finally, to find the angular acceleration when rotating about axis 2, we can rearrange the equation and solve for ω2:
ω2 = (3.6 rad/s^2 * (I_A + I_B)) / (mB * (2A)^2).
Note: To get the final result for the angular acceleration, you need to substitute the values of I_A, I_B, mB, and A into the equation.