American eels (Anguilla rostrata) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could

The ell is observed to spin at 14 spins per second clockwise, and 10 seconds later it was observed to spin at 8 spins per second counterclockwise. What is the magnitude of the ell’s average angular acceleration during this time?

1. $\mathbf{146}\mathbf{/}\mathbf{10}{\mathbf{\text{\hspace{0.17em}rad/s}}}^{\mathbf{2}}$
2. $\mathbf{6}\mathit{\pi }\mathbf{/}\mathbf{10}{\mathbf{\text{\hspace{0.17em}rad/s}}}^{\mathbf{2}}$
3. $\mathbf{12}\mathit{\pi }\mathbf{/}\mathbf{10}{\mathbf{\text{\hspace{0.17em}rad/s}}}^{\mathbf{2}}$
4. $\mathbf{-}\mathbf{44}\mathit{\pi }\mathbf{/}\mathbf{10}{\mathbf{\text{\hspace{0.17em}rad/s}}}^{\mathbf{2}}$
   

Step by step solution

01

Angular acceleration

The time rate at which angular velocity changes is referred to as angular acceleration. The formula for angular acceleration is given by,

$\alpha \left(t\right)=\frac{d\omega }{dt}$

Where$\alpha$is angular acceleration and $\omega$is angular velocity

02

identification of given data

The ell is observed to spin at 14 spins per second clockwise.

So, $n_1=14$

After 10 second, ell is observed to spin at 10 spins per second counter clockwise.

So, $n_2=14$

03

Finding angular acceleration

Now, angular velocity of the ell is given by

$\begin{array}{c}\omega =2\pi \left(14-\frac{14-\left(-8\right)}{10}t\right)\\ =\pi \left(70-\frac{22}{10}t\right)\text{\hspace{0.17em}rad/s}\end{array}$

Now, angular acceleration is given by

$\begin{array}{c}\alpha \left(t\right)=\frac{d\omega }{dt}\\ =\frac{d}{dt}\pi \left(70-\frac{22}{10}t\right)\text{\hspace{0.17em}rad/s}\\ =-\frac{44\pi }{10}{\text{\hspace{0.17em}rad/s}}^2\end{array}$

So, angular acceleration of ell is$-\frac{44\pi }{10}{\text{\hspace{0.17em}rad/s}}^2$

Hence option (d) is correct.

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