A solide homogeneous cylinder of height `h` and base radius `r` is kept vertically on a conveyer belt moving horizontally with an increasing velocity `v=a+bt^(2)`. If the cylinder is not allowed to slip then the time whem the cylinder is about to topple , will be equal to

Options

A. `(rg)/(bh)`
B. `(2rg)/(bh)`
C. `(2bg)/(rh)`
D. `(rg)/(2bh)`

Explanation

To solve the problem, we need to determine the time at which a solid homogeneous cylinder, placed on a conveyor belt with an increasing velocity, is about to topple. Here’s a step-by-step solution:

### Step 1: Understand the Motion of the Cylinder
The cylinder is on a conveyor belt that moves horizontally with a velocity given by the equation:
\[ v = a + bt^2 \]
where \( a \) and \( b \) are constants, and \( t \) is time.

### Step 2: Determine the Acceleration
To find the acceleration of the cylinder, we differentiate the velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt}(a + bt^2) = 2bt \]
Thus, the acceleration \( a \) of the cylinder is:
\[ a = 2bt \]

### Step 3: Analyze Forces Acting on the Cylinder
When the cylinder is about to topple, we need to consider the forces acting on it:
- The weight of the cylinder \( mg \) acts downward at its center of mass.
- The normal force \( N \) acts upward at the base of the cylinder.
- A pseudo force \( F_{\text{pseudo}} = ma \) acts horizontally in the opposite direction of the acceleration of the belt.

### Step 4: Set Up the Torque Equation
For the cylinder to be in equilibrium just before toppling, the torque about the point of contact (point P) must be balanced. The torque due to the weight of the cylinder and the torque due to the pseudo force must be equal.

The torque due to the weight \( mg \) about point P is:
\[ \tau_{\text{weight}} = mg \cdot r \]
where \( r \) is the radius of the cylinder.

The torque due to the pseudo force \( ma \) about point P is:
\[ \tau_{\text{pseudo}} = ma \cdot \frac{h}{2} \]
where \( \frac{h}{2} \) is the distance from the center of mass to point P.

### Step 5: Set the Torques Equal
Setting the torques equal gives us:
\[ mg \cdot r = ma \cdot \frac{h}{2} \]

### Step 6: Substitute for Acceleration
Substituting \( a = 2bt \) into the torque equation:
\[ mg \cdot r = m(2bt) \cdot \frac{h}{2} \]

### Step 7: Simplify the Equation
Cancel \( m \) from both sides:
\[ gr = 2bt \cdot \frac{h}{2} \]
This simplifies to:
\[ gr = bht \]

### Step 8: Solve for Time \( t \)
Rearranging the equation to solve for \( t \):
\[ t = \frac{gr}{bh} \]

### Final Answer
The time when the cylinder is about to topple is:
\[ t = \frac{gr}{bh} \]