Question

A wheel is rotating about an axis that is in the

Answer 1

Answer:

A) $\alpha=+0.625rad/s^2$ is Positive

B) t2 = 6.4s

C) t1 = 9.6s

D) $\Delta \theta =-16rad$

Explanation:

Let's first calculate the aangular acceleration:

$\alpha=\frac{\omega f-\omega o}{\Delta t} = +0.625rad/s^2$

So, acceleration is positive

Let Δt be divided into 2 time intervals t1 and t2 where t1 is the interval when the wheel goes from -6.00 rad/s to 0.00 rad/s

$t1 = \frac{\omega 1 - \omega o}{\alpha}=\frac{0rad/s - (-6rad/s)}{0.625rad/s^2}$

t1 = 9.6s   This is the interval where speed is decreasing

The other time interval t2 will be:

t2 = 16s - t1 = 6.4s  This is the interval where speed is increasing

The angular displacement is:

$\Delta \theta=\omega o *\Delta t+\alpha*\Delta t^2/2$

$\Delta \theta =-16rad$