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Question: A disc rotates about a fixed axis. Its angular velocity \(\omega \) varies with time according to th...

A disc rotates about a fixed axis. Its angular velocity ω\omega varies with time according to the equation ω=at+b\omega = at + b. Initially at t=0t = 0 its angular velocity is 1.0rad/s1.0rad/s and angular position is 2rad2rad; at the instant t=2st = 2s angular velocity is 5.0rad/s5.0rad/s . Determine angular position θ\theta and angular acceleration when t=4st = 4s.

Explanation

Solution

The angle in radians through which a point or line has been rotated about an axis is called its angular position. It is a vector quantity. Angular velocity is defined as how fast the object rotates or changes its position with time. It is represented by using the symbol omega ω\omega .

Complete step-by-step answer:
Step I:
Given that
angular velocityω\omega varies with time according to equation, ω=at+b\omega = at + b ---(i)

When t = 00, angular velocity ω=0.1rad/s\omega = 0.1rad/s
Angular position θ=2πrad\theta = 2\pi rad
Substituting the values in equation(i),
0.1=a×0+b0.1 = a \times 0 + b
b=0.1b = 0.1
Substitute value of ‘b’ in equation (i),
ω=at+0.1\omega = at + 0.1 ---(ii)
Step II:
Angular velocity is given by
ω=dθdt\omega = \dfrac{{d\theta }}{{dt}}
dθ=ω.dtd\theta = \omega .dt
dθ=ω.dt\int {d\theta = \int {\omega .dt} }
Integrating the above equation,
0θdθ=0t(at+0.1)dt\int\limits_0^\theta {d\theta = \int\limits_0^t {(at + 0.1)dt} }
[θ]02=at22+0.1t+2[\theta ]_0^2 = \dfrac{{a{t^2}}}{2} + 0.1t + 2
Step III:
Also given when t=2sect = 2\sec
ω=5rad/s\omega = 5rad/s
Substituting these values in equation(ii)
5=2a+0.15 = 2a + 0.1
2a=50.12a = 5 - 0.1
a=4.92a = \dfrac{{4.9}}{2}
a=2.45a = 2.45
Step IV:
Substituting the value of a in equation (ii)
ω=2.45t+0.1\omega = 2.45t + 0.1
Angular acceleration is the rate of change of the angular velocity and it is a vector quantity. It is given by, =dωdt \propto = \dfrac{{d\omega }}{{dt}}
Substituting value of ω\omega and solving for angular acceleration
=d2.45t+0.1dt\propto = \dfrac{{d\\{ 2.45t + 0.1\\} }}{{dt}}
=2.45rad/s2\propto = 2.45rad/{s^2}
Step V:
Angular position at t=4st = 4s
θ=(2.45)×(4)22+0.1×4+2\theta = (2.45) \times \dfrac{{{{(4)}^2}}}{2} + 0.1 \times 4 + 2
θ=19.6+0.4+2\theta = 19.6 + 0.4 + 2
θ=22rad\theta = 22rad
Step VI:
Therefore, the when t=4st = 4s
Angular position of the disc is θ=22rad\theta = 22rad and
Angular acceleration of the disc is =2.45rad/s2 \propto = 2.45rad/{s^2}

Note: Sometimes there can be confusion between angular frequency and velocity. It is important to note that angular velocity and angular frequency are the same terms. Angular frequency is the magnitude of the angular velocity. It is therefore sometimes also known as the angular velocity. Angular velocity is the product of frequency and the constant 2π2\pi . Whenever any object makes one complete revolution in one second, the object is said to have rotated to a measure of 2π2\pi radians per second.