Question

A rigid body rotates about a stationary axis according to the equation θ = 12t – 4t^3. Its angular acceleration, when its angular velocity becomes zero, is (θ is angular displacement and t is time)​

Answer 1

Solution:

All angular components are analogous to those of linear components. The relation between them is expressed as:

• Displacement = Theta
• Velocity = Angular Velocity
• Acceleration = Angular Acceleration

Angular velocity = d0/dt,

where 0 represents theta.

Angular Acceleration = dw/dt

where w represents angular velocity.

Refer to the image for solution.

Hope it helped!

Answer 2

GIVEN:

$\theta = 12t - 4 {t}^{3}$

TO FIND:

Angular acceleration when angular velocity is zero

ANSWER:

$\alpha \: = 24 \: rad/ {s}^{2}$

EXPLANATION:

As we know,

$\omega = \frac{d \theta}{dt}$

Therefore angular velocity can be calculated as follows;

$= > \omega \: = \frac{d(12t - 4 {t}^{3} )}{dt} \\ \\ = > \omega = \frac{d \: 12t}{dt} - \frac{d \: 4 {t}^{3} }{dt} \\ \\ = > \omega = 12 - 12 {t}^{2} \: rad/ s$

Time when angular velocity is zero;

$= > 12 - 12 {t}^{2} = 0 \\ \\ = > 12 {t}^{2} = 12 \\ \\ = > {t}^{2} = 1 \\ \\ = > t = 1 \: \: (time \: can \: never \: be \: negative)$

Angular acceleration:

$= > \alpha = \frac{d \omega}{dt} \\ \\ = > \alpha = \frac{d (12 - 12 {t}^{2} )}{dt} \\ \\ = > \alpha = \frac{d \: 12}{dt} - \frac{d \: 12 {t}^{2} }{dt} \\ \\ = > \alpha = 24t\: rad/ {s}^{2}$

So, angular acceleration when angular velocity is zero will be;

$= > \alpha \: = 24(1) \\ \\ = > \alpha \: = 24 \: rad/ {s}^{2}$