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WINSTONCH [101]
3 years ago
7

When the person is skating from the bottom of the track back up to the top, her potential energy _______ . When the person is sk

ating from the top of the track down to the bottom, his kinetic energy _______ . When the person is skating from the top of the track down to the bottom, her velocity _______ .
Physics
1 answer:
Alina [70]3 years ago
8 0

Answer:

Explanation:

Potential Energy of an object is a function of elevation from datum thus assuming datum to be ground, the potential energy of the person increases as he moves up to the top.

When a person is skating from the top of the track to the bottom then his potential energy is getting converted into the kinetic energy so his kinetic energy increases as he moves down.

When a person is moving from top to bottom his kinetic energy increases which is a function of mass and velocity. As mass is fixed therefore his velocity increases as he moves down

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At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
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Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

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Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

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