What is your acceleration while sitting in your chair. The latitude of Corvallis

Question

What is your acceleration while sitting in your chair. The latitude of Corvallis is 44.4 degree. What is percent difference in the magnitude of your apparent weight due to the rotational motion? How long would a day have to be for the acceleration to be equal in magnitude to 9.8 m/s^2? The wheel in the figure to the right has eight equally spaced spokes and a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5 rev/s. You want to shoot a 20-cm-long arrow parallel to this axle and through the wheel without hitting any of the spokes. Assume that the arrow and spokes are very thin. What minimum speed must the arrow have? Does it matter where between the axle and the rim of the wheel you aim? If so, what is the best location? An electrical motor spins at 1000 rpm and is slowing down at a rate of 10 rad/s^2. If the motor radius is 7.165 cm, what is the linear acceleration of the edge of the motor? How long will it take to decrease it's angular velocity by 75%?

Explanation / Answer

(a)

formula for accleration due to gravity at a latitude theta is given by

a= w^2 R cos theta

= 6.4 * 10^6 ( 2pi/ 24 ( 3600 s))^2 cos 44.4

=0.02407 m/s^2

(b)

acutal weight is mg

apparent weight due to rotational = mg- mR w^2

percent difference = mRw^2/mg * 100

= Rw^2/g * 100

= 6.4 * 10^6 ( 2pi/ 24 ( 3600 s))^2/9.8 * 100

=3.4 percent

(c)

a= R w^2

9.8= R w^2

9.8 = 6.4 * 10^6 ( 2pi/T)^2

T = 2pi sqrt 6.4 * 10^6/9.8

= 1.4 hours

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