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 7.5

Question

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?

in progress 0
Tài Đức 2 weeks 2021-07-19T14:00:04+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-07-19T14:01:56+00:00

    Answer:

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

    b) a=30.7

    c) a=35.91

    Explanation:

    From the question we are told that:

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

    Initial Length L_1=0.600m

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

    Final Length L_2=0.900m

    a)

    Generally the rotation with the greater speed is

    \omega _1=7.56rev/s=>47.5rads/s

    b)

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

    a=\omega_1^2*L_1

    a=8.16 rev/s*0.6

    a=30.7

    c)

    At 6.35 rev/s

    a=6.35 rev/s*0.9

    a=35.91

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )