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Talja [164]
3 years ago
6

At a certain instant a rigid wheel is spinning about its center of mass with angular velocity of magnitude ω and angular acceler

ation of magnitude α. Consider a point a distance r from the axis of rotation. What would be the magnitude of the acceleration of this point on the wheel?
Physics
1 answer:
geniusboy [140]3 years ago
7 0

Answer:

a = r√(w⁴ + (alpha)²)

Explanation:

let w be the magnitude of angular velocity

r = radius

v = velocity = wr

angular acceleration, ā = v²/r = (wr)²/r = w²r

Also,

â = dv/dt = rdw/dt

given : alpha = dw/dt

hence, â = (alpha)r

the resultant acceleration, a, by Pythagoras is given as

a = √( â² + ā²) = √(r²w⁴ + r²(alpha)²)

simplifying,

a = √r²(w⁴ + (alpha)²)

a = r√(w⁴ + (alpha)²)

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A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of
Dafna11 [192]
In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
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In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
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Now plug in the information we have:
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The first step is to multiply 2 times 8,848m:
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Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
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Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

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