Question

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t)=2t^3-6t^2$ 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

Answer 1

The Correct Option is A

To determine at which time the torque on a rotating wheel becomes zero, we start with the given function for the angular position of a point on the wheel:

The angular position $\theta(t) = 2t^3 - 6t^2$.

The angular velocity $\omega(t)$ is the first derivative of angular position with respect to time:

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

Calculating the derivative:

$\omega(t) = 6t^2 - 12t$

The angular acceleration $\alpha(t)$ is the derivative of angular velocity:

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

Calculating the derivative:

$\alpha(t) = 12t - 12$

The torque $\tau$ is related to the angular acceleration by the equation:

$\tau = I \alpha$, where $I$ is the moment of inertia.

Since we are looking for when the torque is zero, we need:

$0 = I(12t - 12)$

For the torque to be zero, the angular acceleration must be zero:

$12t - 12 = 0$

Solve for $t$:

$12t = 12$

$t = 1$ second.

Hence, the torque on the wheel becomes zero at t = 1 s.