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3 years ago

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A uniform rod is set up so that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod

are 0.899 m0.899 m and 1.13 kg1.13 kg , respectively. A force of constant magnitude FF acts on the rod at the end opposite the rotation axis. The direction of the force is perpendicular to both the rod's length and the rotation axis. Calculate the value of FF that will accelerate the rod from rest to an angular speed of 6.61 rad/s6.61 rad/s in 8.97 s8.97 s .

1 answer:

allsm [11]3 years ago

4 0

Answer:

Explanation:

angular acceleration α = 6.61 / 8.97 = .737 rad / s²

moment of inertia = 1/3 m L²

= 1/3 x 1.13 x .899²

= .3045 kg m²

torque applied = moment of inertia x angular acceleration

= .737 x .3045 = .2244

If be the force

Torque

= F x .899 = .2244

F = .25 N ,

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The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it

Schach [20]

Answer:

The tub of the washer does 37.5 revolutions while it is in motion.

Explanation:

At first we assume that the tub accelerates and decelerates uniformly. In this case, it is required to determine the number of revolutions done by the tub by means of the following equations of motion:

Acceleration

$\ddot n_{1} = \frac{\dot n_{1}-\dot n_{1,o}}{t_{1}}$ (Eq. 1)

Where:

$\ddot n_{1}$ - Angular acceleration experimented by the tub, measured in revolutions per square second.

$\dot n_{1,o}$ , $\dot n_{1}$ - Initial and final angular velocities of the tub, measured in revolutions per second.

$t_{1}$ - Acceleration time, measured in seconds.

$\Delta n_{1} = \dot n_{1,o}\cdot t_{1}+\frac{1}{2}\cdot \ddot n_{1}\cdot t_{1}^{2}$ (Eq. 2)

Where $\Delta n_{1}$ is the change in angular position of the tub during acceleration, measured in revolutions.

Deceleration

$\ddot n_{2} = \frac{\dot n_{2}-\dot n_{1}}{t_{2}}$ (Eq. 3)

Where:

$\ddot n_{2}$ - Angular acceleration experimented by the tub, measured in revolutions per square second.

$\dot n_{1}$ , $\dot n_{2}$ - Initial and final angular velocities of the tub, measured in revolutions per second.

$t_{2}$ - Deceleration time, measured in seconds.

$\Delta n_{2} = \dot n_{1}\cdot t_{2}+\frac{1}{2}\cdot \ddot n_{2}\cdot t_{2}^{2}$ (Eq. 4)

If we know that $\dot n_{1,o} =0\,\frac{rev}{s}$ , $\dot n_{1} = 5\,\frac{rev}{s}$ , $t_{1} = 7\,s$ , $\dot n_{2} = 0\,\frac{rev}{s}$ and $t_{2} = 8\,s$ , then the amount of revolutions done by the tub during each phase is, respectively:

(Eq. 1)

$\ddot n_{1} = \frac{5\,\frac{rev}{s}-0\,\frac{rev}{s} }{7\,s}$

$\ddot n_{1} = \frac{5}{7}\,\frac{rev}{s^{2}}$

(Eq. 3)

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-5\,\frac{rev}{s} }{8\,s}$

$\ddot n_{2} = -\frac{5}{8}\,\frac{rev}{s^{2}}$

(Eq. 2)

$\Delta n_{1} = \left(0\,\frac{rev}{s} \right)\cdot (7\,s)+\frac{1}{2}\cdot \left(\frac{5}{7}\,\frac{rev}{s^{2}}\right) \cdot (7\,s)^{2}$

$\Delta n_{1} = \frac{35}{2}\,rev$

(Eq. 4)

$\Delta n_{2} = \left(5\,\frac{rev}{s} \right)\cdot (8\,s)+\frac{1}{2}\cdot \left(-\frac{5}{8}\,\frac{rev}{s^{2}}\right) \cdot (8\,s)^{2}$

$\Delta n_{2} = 20\,rev$

The total amount of revolutions done by the tub while it is in motion is:

$\Delta n_{T} = \Delta n_{1}+\Delta n_{2}$

$\Delta n_{T} = 37.5\,rev$

The tub of the washer does 37.5 revolutions while it is in motion.

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