A disk at rest experiences constant angular acceleration for t = 25 s, at the end of which it is spinning with a frequency of f = 55 rpm. Please answer the following questions.

A)Write a general expression for the magnitude of the angular velocity of the disk after the acceleration in terms of frequency.

B)Calculate the magnitude of the angular velocity of the disk in rad/s at time t = 25 s.

C)Write a general expression for the average angular acceleration of the disk αave in terms of angular velocity.

D) Calculate the magnitude of the average angular acceleration of the disk αave from rest to t = 25 s in rad/s2.

E)Write a general expression for the period T of the disk in terms of frequency f.

F)Calculate the magnitude of the period of the disk T at time t = 25 s in seconds.

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2022-03-21T11:47:07+01:00

(a) A general expression for the magnitude of the angular velocity of the disk is 2πf.

(b) The magnitude of the angular velocity of the disk is 5.76 rad/s.

(c) A general expression for the average angular acceleration of the disk is ω/t.

(d) The magnitude of the average angular acceleration of the disk is 0.23 rad/s².

(e) A general expression for the period T of the disk is 1/f.

(f) The magnitude of the period of the disk is 0.17 s.

The angular velocity of the disk is the rate of change of angular displacement with time.

ω = 2πf

where;

f is the frequency

[tex]\omega = 2\pi \times 55\frac{rev}{\min} \times \frac{1 \min}{60 \ s} = 5.76 \ rad/s[/tex]

The angular velocity of the disk at time, t = 25 s is 5.76 rad/s.

[tex]\alpha = \frac{\Delta \omega }{t} \\\\\alpha = \frac{5.76}{25} \\\\\alpha = 0.23 \ rad/s^2[/tex]

T = 1/f

T = 1/5.76

T = 0.17 s

Thus, the period of the disk at time t = 25 s is 0.17 s.

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