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

Answer

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 (\( \tau \)) is given by the equation:
\[ \tau = I \cdot \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration. For the torque to be zero, we need \( \alpha = 0 \) since the moment of inertia \( I \) cannot be zero.

### Step 2: Find the angular acceleration
The angular acceleration \( \alpha \) is the second derivative of the angular position \( \theta(t) \) with respect to time \( t \):
\[ \alpha = \frac{d^2\theta}{dt^2} \]
Given the angular position:
\[ \theta(t) = 2t^3 - 6t^2 \]

### Step 3: Differentiate the angular position to find the first derivative (angular velocity)
First, we differentiate \( \theta(t) \) with respect to time \( t \):
\[ \frac{d\theta}{dt} = \frac{d}{dt}(2t^3 - 6t^2) = 6t^2 - 12t \]

### Step 4: Differentiate again to find the angular acceleration
Now, we differentiate the angular velocity to find the angular acceleration:
\[ \alpha = \frac{d^2\theta}{dt^2} = \frac{d}{dt}(6t^2 - 12t) = 12t - 12 \]

### Step 5: Set the 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 \( t \)
Now, we solve for \( t \):
\[ 12t = 12 \implies t = 1 \text{ second} \]

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

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