The instantaneous angular position of a point on a rotating wheel is given by the equation
`theta(t) = 2t^(3) - 6 t^(2)`
The torque on the wheel becomes zero at a) t = 1 s b) t = 0.5 s c) t = 0.25 s d) t = 2 s

Options

A. t = 1 s
B. t = 0.5 s
C. t = 0.25 s
D. t = 2 s

Explanation

To find the time at which the torque on the wheel becomes zero, we start with the given equation for the instantaneous angular position:

\[ \theta(t) = 2t^3 - 6t^2 \]

### Step 1: Find Angular Velocity
First, we need to find the angular velocity \(\omega\), which is the first derivative of \(\theta(t)\) with respect to time \(t\):

\[ \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(2t^3 - 6t^2) \]

Calculating the derivative:

\[ \omega(t) = 6t^2 - 12t \]

### Step 2: Find Angular Acceleration
Next, we find the angular acceleration \(\alpha\), which is the derivative of angular velocity \(\omega(t)\):

\[ \alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt}(6t^2 - 12t) \]

Calculating the derivative:

\[ \alpha(t) = 12t - 12 \]

### Step 3: Set Angular Acceleration to Zero
The torque on the wheel becomes zero when the angular acceleration \(\alpha\) is zero. Therefore, we set \(\alpha(t)\) to zero:

\[ 12t - 12 = 0 \]

### Step 4: Solve for \(t\)
Now, we solve for \(t\):

\[ 12t = 12 \\ t = 1 \text{ s} \]

### Conclusion
Thus, the torque on the wheel becomes zero at:

\[ \boxed{1 \text{ s}} \]

The correct answer is option (a) \(t = 1 \text{ s}\).