A long, thin rod is cut into two pieces, one being twice as long as the other. To the midpoint of piece A (the longer piece), piece B is attached perpendicularly, in order to form the inverted “T” shown in the drawing. The application of a net external torque causes this object to rotate about axis 1 with an angular acceleration of 1.1 rad/s2. When the same net external torque is used to cause the object to rotate about axis 2, what is the angular acceleration?
Well, let me calculate that for you in the most comedic way possible. *Takes out clown calculator*
Alright, let's gather our information here, shall we? We have a long, thin rod that's been chopped into two pieces, one being twice as long as the other. And we're making some fancy inverted "T" construction out of it.
Now, when we apply a net external torque to make this creation rotate about axis 1, we get an angular acceleration of 1.1 rad/s^2. Impressive!
But what happens when we switch things up and rotate it about axis 2? Ah, the suspense! *Drumroll, please*
Well, since we are dealing with a long, thin rod and the lengths of the pieces matter, I must perform some clown-calculations here.
After some clown-math magic, I can tell you that the angular acceleration when rotating about axis 2 will be... *trumpets blaring*... half of the angular acceleration when rotating about axis 1!
So, in this case, the angular acceleration when rotating about axis 2 will be 0.55 rad/s^2. Ta-da!
Remember, folks, always trust a clown with your physics calculations. It's scientifically proven to be extra hilarious!
To answer this question, we need to analyze the rotational dynamics of the object.
Let's consider the initial object before any rotation. We have a long, thin rod cut into two pieces. Let's denote the length of the longer piece as L and the length of the shorter piece as L/2.
When the net external torque is applied, the object will rotate about axis 1. The rotational inertia (I1) of the object about axis 1 can be calculated using the parallel axis theorem.
The parallel axis theorem states that the rotational inertia of an object about an axis parallel to and offset from the axis through its center of mass is equal to the sum of the rotational inertia about the center of mass and the product of the mass and the square of the distance between the two axes.
For the longer piece, the rotational inertia (I_long) about its center of mass can be calculated using the formula for a rod rotating about its center: I = (1/12) * m * L^2.
For the shorter piece, the rotational inertia (I_short) about its center of mass can also be calculated using the same formula: I = (1/12) * m * (L/2)^2.
Since the object rotates about axis 1, the distance between axis 1 and the center of mass of the shorter piece is (L/4).
Using the parallel axis theorem, the total rotational inertia (I1) about axis 1 is given by:
I1 = I_long + I_short + m * (L/4)^2
To find the angular acceleration (α1) about axis 1, we can use Newton's second law for rotational motion:
τ1 = I1 * α1
Given that τ1 = the applied net external torque and α1 = 1.1 rad/s^2, we can solve for I1.
Now, when the same net external torque is used to cause the object to rotate about axis 2, the angular acceleration (α2) can be found using the same equation:
τ2 = I2 * α2
To find I2, we need to calculate the rotational inertia of the object about axis 2. Since axis 2 is perpendicular to axis 1 and passes through the midpoint of piece A, the rotational inertia (I2) is given by:
I2 = I_long + I_short
Now, we can substitute the known values into the equations to find the answers.
To find the angular acceleration when the object is rotated about axis 2, we need to analyze the moment of inertia of the system with respect to axis 2.
The moment of inertia (I) is a measure of an object's resistance to rotational motion and is given by the formula:
I = Σmr²
Where Σ represents the sum, m is the mass of each component, and r is the perpendicular distance of each component from the axis of rotation.
In this case, let's consider piece A and piece B separately.
For piece A:
Since piece A is a long rod, its moment of inertia with respect to either axis 1 or 2 can be represented by the formula for a slender rod rotating about one of its ends:
I_A = (1/3)mL²
Where m is the mass of piece A and L is the length of piece A.
For piece B:
Piece B is attached at the midpoint of piece A, so its perpendicular distance from axis 2 is L/4 (half of the length of piece A). The moment of inertia of piece B with respect to axis 2 is given by:
I_B = mL²/4
Now, we can sum up the total moment of inertia of the system with respect to axis 2:
I_total = I_A + I_B
Substituting the values, we have:
I_total = (1/3)mL² + mL²/4
To simplify the expression, let's find a common denominator:
I_total = (4/12)mL² + (3/12)mL²
I_total = (7/12)mL²
Now we know the moment of inertia with respect to axis 2. Since the net external torque applied to the system is the same, we can use the rotational analog of Newton's second law:
τ = Iα
Where τ is the net external torque and α is the angular acceleration.
Since the torque is the same for both rotations, we can equate the two expressions:
τ = I₁α₁ = I₂α₂
Where I₁ and α₁ represent the moment of inertia and angular acceleration when rotating about axis 1, and I₂ and α₂ represent the moment of inertia and angular acceleration when rotating about axis 2.
Rearranging the equation, we can solve for α₂:
α₂ = (I₁/I₂)α₁
Substituting the moment of inertias we found earlier, we have:
α₂ = [(1/3)mL²/((7/12)mL²)]α₁
α₂ = (4/7)α₁
Now we know that the angular acceleration about axis 2 is 4/7 times the angular acceleration about axis 1.
Therefore, the angular acceleration when the same net external torque is used to cause the object to rotate about axis 2 is 4/7 times the angular acceleration when rotating about axis 1, or (4/7)(1.1 rad/s²).
Obviously I do not know what your axes are.
lengths 2 and 1
masses 2 and 1
if along the long rod:
I = (1/3)(1) * 1^2 = 1/3
if along the short rod
I = (1/12)(2) * 2^2 = 8/12 = 2/3
if perpendicular to the plane of the rods at intersection
I = 3 (1/3) (1)(1^2) = 1 = 3/3
so angular accelerations are in the ratio 1,2,3 with the maximum being along the long rod through its center and the minimum (max I) being the last of the three