Question

When the gear rotates 20 revolutions, it achieves an angular velocity of v = 30 rad>s, starting from rest. Determine its constant angular acceleration and the time required?

Answer 1

Step-by-step explanation:

It is given that,

Displacement,
`\theta=2\ rev=12.56\ rad`

Initial angular velocity,
`\omega_i=0`

Final angular velocity,
`\omega_f=30\ rad/s`

Let
`\alpha` is its angular acceleration. Using the third equation of kinematics to find it as :

`\omega_f^2-\omega_i^2=2\alpha \theta`

`\omega_f^2=2\alpha \theta`

`\alpha =(\omega_f^2)/(2\theta)`

`\alpha =((30)^2)/(2* 12.56)`

Angular acceleration,
`\alpha =35.82\ rad/s^2`

Let t is the time required to find it as :

`t=(\omega_f-\omega_i)/(\alpha )`

`t=(30)/(35.82)`

t = 0.83 seconds

So, the angular acceleration of the gear is
`35.82\ rad/s^2` and the time taken is 0.83 seconds.

Answer 2

solution;

`final angular velocity w_(f)=30ras/s\\</p><p>initial w_(o)=0\\</p><p>w_(f)^(2)-w_(0)^(2)=2\alpha \theta \\</p><p>(30)^2=2.\alpha .20*2\pi \\</p><p>\alpha =((30)^2)/(2*2\pi *20)\\</p><p>\alpha =3.58rad/s^2\\</p><p>w_(f)=w_(o)+\alpha t\\</p><p>t=(wf)/(\alpha )=(30)/(3.58)\\</p><p>t=8.38`