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Temka [501]
3 years ago
7

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.6 mm.
If the thrower takes 1.0 s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Physics
1 answer:
yawa3891 [41]3 years ago
6 0

Answer:

Explanation:

Time taken to complete one revolution is called time period.

So, Time period, T = 1 s

Diameter = 1.6 mm

radius, r = 0.8 mm

Let the angular speed is ω.

The relation between angular velocity and the time period is

\omega =\frac{2\pi}{T}

ω = 2 x 3.14 = 6.28 rad/s

The relation between the linear velocity and the angular velocity is

v = r x ω

v = 0.8 x 10^-3 x 6.28

v = 0.005 m/s

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Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

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If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

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b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

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x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

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