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

(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.

Step-by-step 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)=(L_(f)-L_(i))/(t)`

Put the value into the formula

`\tau_(avg)=(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+(\alpha t^2)/(2)`

Here,
`\alpha=(\tau)/(I)`

`\omega=(L_(i))/(I)`

Put the value of angular acceleration and angular velocity

`\theta=(L_(i)t)/(I)+(\tau t^2)/(2I)`

`\theta=(L_(i)t+\tau* (t^2)/(2))/(I)`

Put the value into the formula

`\theta=(3.00*1.50-(1.47*1.50^2)/(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*20.3`

`W=-29.84\ J`

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

Using formula of average power

`P=-(W)/(\Delta t)`

Put the value into the formula

`P=-(-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.

Answer 2

given,

I = 0.140 kg ·m²

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

a)
`\tau = (\Delta L)/(\Delta t)`

`\tau = (0.8-3)/(1.5)`

τ = -1.467 N m

b) angle at which fly wheel will turn

`\theta= \omega t +(1)/(2)\alpha t^2`

`\theta= (L)/(I) t +(1)/(2)(\tau)/(I)t^2`

`\theta= (3)/(0.14)* 1.5+(1)/(2)(-1.467)/(0.14)* 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) =-(W)/(t)`

`P_(av) =-((-29.86))/(1.5)`

`P_(av) =19.91\ W`