Question

asked Mar 6, 2021 133k views

1 Answer

Fodon · Answer 1 · 2021-03-10T11:20:57+0000

Answer:

a)
$\Delta \theta_(T) = 527.22\,rad$ , b)
$t = 12.493\,s$ , c)
$\alpha_(2) = -7.556\,(rad)/(s^(2))$

Step-by-step explanation:

a) The travelled angular position during acceleration is:

$\Delta \theta_(1) = (25\,(rad)/(s))\cdot (1.80\,s) +(1)/(2)\cdot (31\,(rad)/(s^(2)) )\cdot (1.80\,s)^(2)$

$\Delta \theta_(1) = 95.22\,rad$

The travelled angular position during deceleration is:

$\Delta \theta_(2) = 432\,rad$

The total travelled angular position is:

$\Delta \theta_(T) = \Delta \theta_(1) + \Delta \theta_(2)$

$\Delta \theta_(T) = 527.22\,rad$

b) The final angular speed at acceleration phase is:

$\omega^(2) = \omega_(o)^(2) + 2\cdot \alpha_(1)\cdot \Delta \theta_(1)$

$\omega = \sqrt{\omega_(o)^(2)+2\cdot \alpha_(1)\cdot \Delta \theta_(1)}$

$\omega = \sqrt{(25\,(rad)/(s) )^(2)+2\cdot (31\,(rad)/(s^(2)) )\cdot (95.22\,rad)}$

$\omega = 80.8\,(rad)/(s)$

The angular deceleration experimented by the grinding wheel is:

$\omega^(2) = \omega_(o)^(2) + 2\cdot \alpha_(2)\cdot \Delta \theta_(2)$

$\alpha_(2) = (\omega^(2)-\omega_(o)^(2))/(2\cdot \Delta \theta_(2))$

$\alpha_(2) = ((0\,(m)/(s) )^(2)-(80.8\,(m)/(s) )^(2))/(2\cdot (432\,rad))$

$\alpha_(2) = -7.556\,(rad)/(s^(2))$

The period of time of the deceleration phase is:

$\omega = \omega_(o) +\alpha_(2)\cdot \Delta t_(2)$

$\Delta t_(2) = (\omega - \omega_(o))/(\alpha_(2))$

$\Delta t_(2) = (0\,(rad)/(s)-80.8\,(rad)/(s))/(-7.556\,(rad)/(s^(2)) )$

$\Delta t_(2) = 10.693\,s$

The instant when the grinding wheel stops is:

$t = 1.80\,s + 10.693\,s$

$t = 12.493\,s$

c) The angular deceleration is:

$\alpha_(2) = -7.556\,(rad)/(s^(2))$

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