To find the dimension of angular momentum in a system where energy (E), momentum (P), and force (F) are chosen as fundamental units, we can follow these steps:
### Step 1: Define Angular Momentum
Angular momentum (L) is defined as the product of the moment of inertia and angular velocity. Mathematically, it can be expressed as:
\[ L = mvr \]
where:
- \( m \) = mass
- \( v \) = linear velocity
- \( r \) = radius (or distance from the axis of rotation)
### Step 2: Write the Dimensions of Angular Momentum
The dimensions of angular momentum can be expressed in terms of fundamental dimensions:
- Mass (M) has the dimension [M]
- Velocity (v) has the dimension [L T^{-1}]
- Radius (r) has the dimension [L]
Thus, the dimensions of angular momentum can be written as:
\[ [L] = [M][L][L T^{-1}] = [M][L^2][T^{-1}] \]
### Step 3: Express Angular Momentum in Terms of E, P, and F
We assume that angular momentum (L) can be expressed as a combination of the fundamental units E, P, and F:
\[ L = E^a P^b F^c \]
### Step 4: Write the Dimensions of E, P, and F
Now, we need the dimensions of energy (E), momentum (P), and force (F):
1. **Energy (E)**:
- Energy is defined as work done, which is force times distance.
- Dimensions of force (F) = [M L T^{-2}]
- Thus, dimensions of energy = [F][L] = [M L^2 T^{-2}].
2. **Momentum (P)**:
- Momentum is defined as mass times velocity.
- Dimensions of momentum = [M][L T^{-1}] = [M L T^{-1}].
3. **Force (F)**:
- Dimensions of force = [M L T^{-2}].
### Step 5: Substitute the Dimensions into the Equation
Now we can substitute the dimensions of E, P, and F into our expression for angular momentum:
\[ [L] = [E^a][P^b][F^c] \]
This gives us:
\[ [M^a L^{2a} T^{-2a}][M^b L^b T^{-b}][M^c L^{c} T^{-2c}] \]
### Step 6: Combine the Dimensions
Combining the dimensions, we get:
\[ [M^{a+b+c} L^{2a+b+c} T^{-2a-b-2c}] \]
### Step 7: Set Up the Equations
Now we can equate the dimensions from both sides:
1. For mass (M):
\[ 1 = a + b + c \] (1)
2. For length (L):
\[ 2 = 2a + b + c \] (2)
3. For time (T):
\[ -1 = -2a - b - 2c \] (3)
### Step 8: Solve the Equations
From equation (1):
\[ c = 1 - a - b \]
Substituting \( c \) into equation (2):
\[ 2 = 2a + b + (1 - a - b) \]
This simplifies to:
\[ 2 = a + 1 \]
Thus, \( a = 1 \).
Now substituting \( a = 1 \) into equation (1):
\[ 1 = 1 + b + c \]
This gives us:
\[ b + c = 0 \]
So, \( c = -b \).
Substituting \( a = 1 \) into equation (3):
\[ -1 = -2(1) - b - 2(-b) \]
This simplifies to:
\[ -1 = -2 - b + 2b \]
Thus:
\[ -1 + 2 = b \]
So, \( b = 1 \) and \( c = -1 \).
### Step 9: Final Expression for Angular Momentum
Now we can express angular momentum in terms of E, P, and F:
\[ L = E^1 P^1 F^{-1} \]
### Conclusion
Thus, the dimension of angular momentum in the new system is:
\[ L = E P F^{-1} \]
