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

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 need to follow these steps:

### Step 1: Understand the relationship between torque and angular acceleration
Torque (τ) is given by the equation:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration. For the torque to be zero, either \( I \) must be zero (which is not the case here since it is a constant non-zero value) or \( \alpha \) must be zero.

### Step 2: Find the angular position function
The angular position \( \theta(t) \) is given by:
\[ \theta(t) = 2t^3 - 6t^2 \]

### Step 3: Calculate angular velocity
To find the angular acceleration, we first need to find the angular velocity \( \omega \), which is the derivative of angular position with respect to time:
\[ \omega(t) = \frac{d\theta}{dt} \]
Calculating the derivative:
\[ \omega(t) = \frac{d}{dt}(2t^3 - 6t^2) \]
Using the power rule:
\[ \omega(t) = 6t^2 - 12t \]

### Step 4: Calculate angular acceleration
Next, we find the angular acceleration \( \alpha \), which is the derivative of angular velocity with respect to time:
\[ \alpha(t) = \frac{d\omega}{dt} \]
Calculating the derivative:
\[ \alpha(t) = \frac{d}{dt}(6t^2 - 12t) \]
Using the power rule again:
\[ \alpha(t) = 12t - 12 \]

### Step 5: Set angular acceleration to zero
To find when the torque is zero, we set the angular acceleration \( \alpha \) to zero:
\[ 12t - 12 = 0 \]

### Step 6: Solve for time \( t \)
Solving the equation:
\[ 12t = 12 \]
\[ t = 1 \text{ second} \]

### Conclusion
Thus, the torque on the wheel becomes zero at \( t = 1 \) second.

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