Respuesta :
Answer
given,
I = 0.140 kg ·m²
decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.
a) [tex]\tau = \dfrac{\Delta L}{\Delta t}[/tex]
[tex]\tau = \dfrac{0.8-3}{1.5}[/tex]
τ = -1.467 N m
b) angle at which fly wheel will turn
[tex]\theta= \omega t +\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2[/tex]
[tex]\theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2[/tex]
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
[tex]P_{av} =-\dfrac{W}{t}[/tex]
[tex]P_{av} =-\dfrac{(-29.86)}{1.5}[/tex]
[tex]P_{av} =19.91\ W[/tex]
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
[tex]\tau_{avg}=\dfrac{L_{f}-L_{i}}{t}[/tex]
Put the value into the formula
[tex]\tau_{avg}=\dfrac{0.800-3.00}{1.50}[/tex]
[tex]\tau_{avg}=-1.47\ N-m[/tex]
[tex]|\tau_{avg}|=1.47\ N-m[/tex]
(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
[tex]\theta=\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]
Here, [tex]\alpha=\dfrac{\tau}{I}[/tex]
[tex]\omega=\dfrac{L_{i}}{I}[/tex]
Put the value of angular acceleration and angular velocity
[tex]\theta=\dfrac{L_{i}t}{I}+\dfrac{\tau t^2}{2I}[/tex]
[tex]\theta=\dfrac{L_{i}t+\tau\times \dfrac{t^2}{2}}{I}[/tex]
Put the value into the formula
[tex]\theta=\dfrac{3.00\times1.50-\dfrac{1.47\times1.50^2}{2}}{0.140}[/tex]
[tex]\theta=20.3\ rad[/tex]
(c). We need to calculate the work done on the wheel
Using formula of work done
[tex]W=F\cdot d[/tex]
[tex]W=Fr\theta[/tex]
[tex]W=\tau\theta[/tex]
Put the value into the formula
[tex]W=-1.47\times20.3[/tex]
[tex]W=-29.84\ J[/tex]
(d). We need to calculate the average power of the flywheel
Using formula of average power
[tex]P=-\dfrac{W}{\Delta t}[/tex]
Put the value into the formula
[tex]P=-\dfrac{-29.84}{1.50}[/tex]
[tex]P=19.89\ W[/tex]
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.
