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3 years ago

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Young David who slew Goliath experimented with slings beforetackling the giant. He found that he could revolve a sling oflength

0.600 m at the rate of9.00 rev/s. If he increased the lengthto 0.900 m, he could revolve the slingonly 7.00 times per second.
(a) What is the speed of the stone for eachrate of rotation?
1 m/s at 9.00 rev/s
2 m/s at 7.00 rev/s

(b) What is the centripetal acceleration of the stone at9.00 rev/s?
3 m/s2

(c) What is the centripetal acceleration at 7.00 rev/s?
4m/s2

1 answer:

Mars2501 [29]3 years ago

3 0

Answer:

a) v1 = 33.9 m/s , v2 = 39.6 m/s, b) a = 1015.4 m/s² and c) a = 1742.4 m/s²

Explanation:

To solve this exercise we will use the relationship between angular and linear quantities

a) The angular and linear velocity are related by

v = w r

Let's reduce the quantities to the SI system

w1 = 9.00 rev / s (2π rad / 1rev) = 18 π rad / s = 56.55 rad / s

w2 = 7.00 rev / s = 14π rad / s = 43.98vrad / s

v1 = 56.55 0.6

v1 = 33.9 m / s

v2 = 43.98 0.9

v2 = 39.6 m / s

b) centripetal acceleration for v1

a = v² / r

a = 33.9² / 0.6

a = 1015.4 m / s²

c) centripetal acceleration for v2

a = 39.6² / 0.9

a = 1742.4 m / s²

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