Question

A discus thrower (Fig. P4.27, page 97) accelerates a discus from rest to a speed of 25.0 m/s by whirling it through 1.25 rev. Assume the discus moves on the arc of a circle 1.00m in radius.(c) Calculate the time interval required for the discus to accelerate from rest to 25.0 m/s

Answer 1

The time interval needed for the disk to rev from leave to 25.0 m/s

= 0.57 s

What is the time interval?

• A larger period of time can be split up into multiple shorter, equal-length segments.
• These are referred to as time periods. Consider the scenario where you wished to gauge a car's speed over an hour-long trip. You could break up an hour into ten-minute segments.
• In Hz, frequency is defined as the number of cycles per second. Simply divide 1 by the frequency to determine the time interval for a known frequency (e.g., a frequency of 100 Hz has a time interval of 1/(100 Hz) = 0.01 seconds; 500 Hz has a time interval of 1/(500Hz) = 0.002 seconds, etc.).

Given:

vo = 0

v = 25.4 m/s

r = 0.95 m

n = 1.21 rev

A.

r*ω = v

ω = v/r

= 25.4/.95

= 26.74 rad/s

B.

θ = 2πn

= 2π × (1.21)

= 7.6 m

Using equation of motion,

ωf^2 = ωi^2 + 2αθ

(26.74)^2 = 2 × α × (7.6)

α = 715.03/15.2

= 47.04 rad/s^2

C.

Using equation of motion,

θ = ωi*t + 1/2*α*t^2

θ = 0 + 1/2*α*t^2

7.6 = 47.04/2 × t^2

= sqrt(0.323)

= 0.57 s

Therefore the correct answer is 0.57 s

To learn more about time interval, refer to:

brainly.com/question/479532

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