Question

The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.800 kg ·m2/s in 1.50 s.
(a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period?
(b) Assuming a constant angular acceleration, through what angle does the flywheel turn?
(c) How much work is done on the wheel?
(d) What is the average power of the flywheel?

Answer 1

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) $\tau = \dfrac{\Delta L}{\Delta t}$

   $\tau = \dfrac{0.8-3}{1.5}$

        τ = -1.467 N m

b) angle at which fly wheel will turn

   $\theta= \omega t +\dfrac{1}{2}\alpha t^2$

   $\theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2$

   $\theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2$

        θ = 20.35 rad

c) work done on the wheel

     W = τ x θ

     W = -1.467 x 20.35 rad

    W = -29.86 J

d) average power of wheel

    $P_{av} =-\dfrac{W}{t}$

    $P_{av} =-\dfrac{(-29.86)}{1.5}$

     $P_{av} =19.91\ W$          

Answer 2

Answer:

(a). The magnitude of the average torque acting on the flywheel about its central axis during this period is 1.47 N-m.

(b). The angle that the flywheel does turn is 20.3 rad.

(c). The work done is -29.84 J.

(d). The average power of the flywheel is 19.89 W.

Explanation:

Given that,

Rotational inertia = 0.140 kg m²

Initial angular momentum of flywheel = 3.00 kg m²/s

Final angular momentum of flywheel = 0.800 kg m²/s

Time = 1.50 sec

(a). We need to calculate the magnitude of the average torque acting on the flywheel about its central axis during this period

Using formula of torque

$\tau_{avg}=\dfrac{L_{f}-L_{i}}{t}$

Put the value into the formula

$\tau_{avg}=\dfrac{0.800-3.00}{1.50}$

$\tau_{avg}=-1.47\ N-m$

$|\tau_{avg}|=1.47\ N-m$

(b). Assuming a constant angular acceleration, through what angle does the flywheel turn

We need to calculate the angle that the flywheel does turn

Using equation for angle

$\theta=\omega_{0}t+\dfrac{\alpha t^2}{2}$

Here, $\alpha=\dfrac{\tau}{I}$

$\omega=\dfrac{L_{i}}{I}$

Put the value of angular acceleration and angular velocity

$\theta=\dfrac{L_{i}t}{I}+\dfrac{\tau t^2}{2I}$

$\theta=\dfrac{L_{i}t+\tau\times \dfrac{t^2}{2}}{I}$

Put the value into the formula

$\theta=\dfrac{3.00\times1.50-\dfrac{1.47\times1.50^2}{2}}{0.140}$

$\theta=20.3\ rad$

(c). We need to calculate the work done on the wheel

Using formula of work done

$W=F\cdot d$

$W=Fr\theta$

$W=\tau\theta$

Put the value into the formula

$W=-1.47\times20.3$

$W=-29.84\ J$

(d). We need to calculate the average power of the flywheel

Using formula of average power

$P=-\dfrac{W}{\Delta t}$

Put the value into the formula

$P=-\dfrac{-29.84}{1.50}$

$P=19.89\ W$

Hence, (a). The magnitude of the average torque acting on the flywheel about its central axis during this period is 1.47 N-m.

(b). The angle that the flywheel does turn is 20.3 rad.

(c). The work done is -29.84 J.

(d). The average power of the flywheel is 19.89 W.