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

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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the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s