Answer :
Answer:
A) Must be done 19806.62 joules of work.
B) The average power is 1320.44 Watts.
Explanation:
A) First, we're going to use the work-energy theorem that states total work ([tex] W[/tex]) done on an object is equal to the change in its kinetic energy ([tex] \Delta K[/tex]):
[tex]W=\Delta K = K_{f}-K_{i}[/tex] (1)
So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation [tex]K=\frac{I\omega^{2}}{2} [/tex] for the kinetic energy so:
[tex] \Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2}[/tex] (2)
with [tex] \omega_{i}[/tex] the initial angular velocity, [tex]\omega_{f} [/tex] the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is [tex] I=MR^{2} [/tex], using those on (2)
[tex] \Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2} [/tex] (3)
By (3) on (1):
[tex]W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2} [/tex]
[tex]W=19806.62\,J [/tex]
B) Average power is work done divided by the time interval:
[tex]P=\frac{W}{\Delta t}=\frac{19806.62}{15.0} [/tex]
[tex] P=1320.44\,W[/tex]
NOTE: We use the relation [tex]1rpm*\frac{2\pi}{60s}=\frac{rad}{s} [/tex] to convert 280 rev/min(rpm) to 29.32 rad/s
A. The amount of work that must be done to stop the wheel from rotating is -19.8 Kilojoules.
B. The required average power: is equal to 1.32 Kilowatts.
Given the following data:
- Mass of wheel = 32.0 kg
- Radius of thin hoop = 1.20 m
- Angular velocity = 280 rev/min.
- Time = 15.0 seconds.
Conversion:
Angular velocity = 280 rev/min to rad/s = [tex]280 \times \frac{2\pi}{60} = \frac{1759.52}{60} = 29.33\; rad/s[/tex]
A. To determine the amount of work that must be done to stop the wheel from rotating, we would apply the work-kinetic energy theorem:
According to the work-kinetic energy theorem, we have:
[tex]W = \Delta K.E\\\\W = K.E_f - K.E_i[/tex] ....eqn. 1.
For rotational motion:
[tex]K.E = \frac{1}{2} I\omega^2[/tex]
Where:
I is the moment of inertia.
- [tex]\omega[/tex] is the angular velocity.
[tex]\Delta K.E = \frac{1}{2} I\omega_f^2 - \frac{1}{2} I\omega_i^2[/tex] ...eqn. 2.
The final angular velocity of the wheel is equal to zero when it comes to a stop.
For a thin hoop, the moment of inertia is given by:
[tex]I = mr^2[/tex] ....eqn. 3.
Substituting eqn. 2 and 3 into eqn. 1, we have:
[tex]W = 0- \frac{1}{2} mr^2\omega_i^2\\\\W = -\frac{m(r\omega_i)^2}{2} \\\\W = -\frac{32 \times(1.2 \times 29.33)^2}{2} \\\\W=-\frac{32 \times (35.2)^2}{ 2}\\\\W=-\frac{32 \times 1239.04}{ 2}\\\\W = -\frac{39649.28}{2}[/tex]
Work, W = -19,824.84 Joules
Work, W = -19.8 Kilojoules.
B. To determine the required average power:
[tex]Power = \frac{Work}{Time} \\\\Power = \frac{19.8}{15}[/tex]
Power = 1.32 Kilowatts.
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