An object will continue moving uniformly until

Options

A. the resultant force acting on it begins to decrease
B. the resultant force on it is zero
C. the resultant force is at right angle to its rotation
D. the resultant force on it is increased continuously

Answer

To solve the problem of finding the ratio of the moment of inertia about an axis perpendicular to a rectangular plate passing through points O and O', we can follow these steps:

### Step 1: Understand the Geometry
We have a rectangular plate with dimensions 60 cm and 80 cm. The points O and O' are located such that O is perpendicular to the plate. We need to find the distance between O and O' using the Pythagorean theorem.

### Step 2: Calculate the Length OP
Using the Pythagorean theorem:
\[ OP = \sqrt{(60 \, \text{cm})^2 + (80 \, \text{cm})^2} \]
Calculating:
\[ OP = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \, \text{cm} \]

### Step 3: Find the Distance OO'
Since O' is the midpoint of OP, we can find OO' as:
\[ OO' = \frac{OP}{2} = \frac{100 \, \text{cm}}{2} = 50 \, \text{cm} \]

### Step 4: Moment of Inertia about O'
The moment of inertia \(I_{O'}\) about point O' can be calculated using the formula:
\[ I_{O'} = \frac{m}{12} (a^2 + b^2) \]
where \(a = 60 \, \text{cm}\) and \(b = 80 \, \text{cm}\). Thus:
\[ I_{O'} = \frac{m}{12} (60^2 + 80^2) = \frac{m}{12} (3600 + 6400) = \frac{m}{12} (10000) = \frac{10000m}{12} \]

### Step 5: Moment of Inertia about O
Using the parallel axis theorem, the moment of inertia about point O can be calculated as:
\[ I_{O} = I_{O'} + m \cdot d^2 \]
where \(d = OO' = 50 \, \text{cm}\):
\[ I_{O} = \frac{10000m}{12} + m \cdot (50^2) = \frac{10000m}{12} + m \cdot 2500 \]
Factoring out \(m\):
\[ I_{O} = m \left(\frac{10000}{12} + 2500\right) \]

### Step 6: Simplify the Expression for \(I_{O}\)
To combine the terms:
\[ I_{O} = m \left(\frac{10000}{12} + \frac{2500 \cdot 12}{12}\right) = m \left(\frac{10000 + 30000}{12}\right) = m \left(\frac{40000}{12}\right) \]

### Step 7: Find the Ratio of \(I_{O}\) to \(I_{O'}\)
Now, we can find the ratio:
\[ \frac{I_{O}}{I_{O'}} = \frac{m \cdot \frac{40000}{12}}{m \cdot \frac{10000}{12}} = \frac{40000}{10000} = 4 \]

### Step 8: Final Ratio
Thus, the ratio of the moment of inertia about O to that about O' is:
\[ \frac{I_{O}}{I_{O'}} = 4 \]