To find the dimensions of mass (M) in terms of time (T), velocity (C), and angular momentum (H), we can follow these steps:
### Step 1: Establish the relationship
Assume that mass (M) can be expressed in terms of the fundamental quantities T, C, and H as follows:
\[ M \propto T^x C^y H^z \]
where \(x\), \(y\), and \(z\) are the powers we need to determine.
### Step 2: Write the dimensions of the quantities
We know the dimensions of the quantities involved:
- Time (T) has dimensions: \([T] = T^1\)
- Velocity (C) has dimensions: \([C] = L^1 T^{-1}\)
- Angular momentum (H) has dimensions: \([H] = M^1 L^2 T^{-1}\)
### Step 3: Write the dimensions of mass
The dimensions of mass (M) are:
\[ [M] = M^1 L^0 T^0 \]
### Step 4: Substitute dimensions into the equation
Substituting the dimensions of T, C, and H into the equation for mass gives us:
\[ M^1 = T^x (L^1 T^{-1})^y (M^1 L^2 T^{-1})^z \]
Expanding this, we have:
\[ M^1 = T^x L^y T^{-y} M^z L^{2z} T^{-z} \]
Combining the terms, we get:
\[ M^1 = M^z L^{y + 2z} T^{x - y - z} \]
### Step 5: Equate the dimensions
Now we can equate the coefficients of M, L, and T from both sides:
1. For M:
\[ z = 1 \]
2. For L:
\[ y + 2z = 0 \]
3. For T:
\[ x - y - z = 0 \]
### Step 6: Solve the equations
From the first equation, we have:
\[ z = 1 \]
Substituting \(z = 1\) into the second equation:
\[ y + 2(1) = 0 \]
\[ y + 2 = 0 \]
\[ y = -2 \]
Now substituting \(y = -2\) and \(z = 1\) into the third equation:
\[ x - (-2) - 1 = 0 \]
\[ x + 2 - 1 = 0 \]
\[ x + 1 = 0 \]
\[ x = -1 \]
### Step 7: Write the final expression for mass
Now we have:
- \(x = -1\)
- \(y = -2\)
- \(z = 1\)
Thus, we can express mass in terms of T, C, and H:
\[ M = T^{-1} C^{-2} H^{1} \]
### Final Answer
The dimensions of mass in terms of time, velocity, and angular momentum are:
\[ M = T^{-1} C^{-2} H^{1} \]
