Question

The tape is now wound back into the drum at angular rate omega(t). With what speed will the end of the tape move? (Note that our drawing specifies that a positive derivative of x(t)implies motion away from the drum. Be careful with your signs! The factthat the tape is being wound back into the drum implies that \omega(t) < 0, and for the end of the tape to move closer to the drum, it must be the case that v(t) < 0.
Answer in terms of omega(t) and other given quantities from the problem introduction.
v(t) =_____.

Answer 1

Answer:

See description

Explanation:

Frist we need to know the longitude of tape which is unwinding. Such relationship can be obtained with arc length. An arc length is the distance bewteen two points in a curve.

The relationship is:

$S = \theta r$

Where $S$ is the arc length or distance, theta is the angle that results from the initial point of the measure to the final point, and r is the radius of a circumference.

Now let $x(t)$ be length the unwinded tape. Change $S$ by $x(t)$ and you ge the relationship:

$x(t) = \theta r$

if you unwind the tape by one revolution ($\theta = 2 \pi$) you get the perimeter of a cricle $x(t)= 2 \pi r$, if you unwind it two times then $x(t)= 4 \pi r$ and so on.

Then we have that the derivative of $x(t)$ is $v(t)$

so we replace:

$\frac{dx}{dt} = v(t)\\ \frac{dx}{dt}=v(t)=\frac{d\theta}{dt}$

the derivative of theta with respect to t is ω(t) by definition:

$\frac{d\theta}{dt}=\omega(t)\\ =>\frac{dx}{dt}=v(t)=\omega(t) r\\=>\frac{v(t)}{r}=\omega(t)$

The result is the relationship between angular velocity and the velocity and tangential velocity at the point r