With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s. What is its angular acceleration in rad/s2? How many revolutions does it go through in the process?

Respuesta :

Angular acceleration is 80 rad/s²

Number of revolutions undergone is 1.02

Explanation:

We have equation of motion v = u + at

     Initial angular velocity, u = 0 rad/s

     Final angular velocity, v = 32 rad/s    

     Time, t = 0.40 s

     Substituting

                      v = u + at  

                      32 = 0 + a x 0.40

                      a = 80 rad/s²

     Angular acceleration is 80 rad/s²

  We have equation of motion s = ut + 0.5 at²

        Initial angular velocity, u = 0 rad/s

        Angular acceleration, a = 80 rad/s²

        Time, t = 0.4 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 0.4 + 0.5 x 80 x 0.4²

                      s = 6.4 rad

      Angular displacement  = 6.4 rad

     [tex]\texttt{Number of revolutions = }\frac{6.4}{2\pi}=1.02[/tex]

Number of revolutions undergone is 1.02

This question involves the concepts of the equations of motion for angular motion.

a) The angular acceleration will be "80 rad/s²".

b) It goes through "1.02 revolutions" in the process.

a)

We will use the first equation of motion for angular motion to find out the angular acceleration:

[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

where,

[tex]\alpha[/tex] = angular acceleration = ?

[tex]\omega_f[/tex] = final angular speed = 32 rad/s

[tex]\omega_i[/tex] = initial angular speed = 0 rad/s

t = time taken = 0.4 s

Therefore,

[tex]\alpha =\frac{32\ rad/s-0\ rad/s}{0.4\ s}\\\\\alpha= 80\ rad/s^2[/tex]

b)

Now, we will use the second equation of motion for the angular motion to find out the no. of revolutions:

[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2[/tex]

where,

θ = angualr displacement = ?

Therefore,

[tex]\theta=(0\ rad/s)(0.4\ s)+\frac{1}{2}(80\ rad/s^2)(0.4\ s)^2\\\\\theta=6.4\ rad[/tex]

Now, the number of revolutions (N) are given as follows:

[tex]N=\theta(\frac{1\ rev}{2\pi\ rad})\\\\N=(6.4\ rad)(\frac{1\ rev}{2\pi\ rad})\\\\[/tex]

N = 1.02 revolutions

Learn more about the angular motion here:

brainly.com/question/14979994?referrer=searchResults

The attached picture shows the angular equations of motion.

Ver imagen hamzaahmeds