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An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.

Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?

Respuesta :

okpalawalter8

Answer:

a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

b) [tex]a=30.7[/tex]

c) [tex]a=35.91[/tex]

Explanation:

From the question we are told that:

Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

Initial Length [tex]L_1=0.600m[/tex]

Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]

Final Length [tex]L_2=0.900m[/tex]

a)

Generally the rotation with the greater speed is

[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

b)

Generally the equation for centripetal acceleration at 8.16 is mathematically given by

[tex]a=\omega_1^2*L_1[/tex]

[tex]a=8.16 rev/s*0.6[/tex]

[tex]a=30.7[/tex]

c)

At 6.35 rev/s

[tex]a=6.35 rev/s*0.9[/tex]

[tex]a=35.91[/tex]