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bezimeni [28]
1 year ago
10

you are sitting at the center of a large turntable at an amusement park as it is set spinning freely. you decide to crawl toward

s the edge of the turntable. what does your action do to the rate of rotation of the turntable? explain
Physics
1 answer:
larisa [96]1 year ago
7 0

You are sitting at the center of a large turntable at an amusement park as it is set spinning freely. You decide to crawl towards the edge of the turntable. Rotational speed will decrease

There is no external torque

hence , the angular momentum of the table is conserved

L (initial) = L ( final)

since , L = m*v*r

where

m = mass

v = velocity

r = radius

If  the person is crawled towards the outer rim, then the rotational inertia of the turntable will increase. In order to conserve the angular momentum , its rotational speed will decrease as mass and radius cannot be altered .

To learn more about torque here :

brainly.com/question/19104073

#SPJ4

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olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

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Since we are looking for the shortest time we will pick t = 9.22 s

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