Question

A rigid body rotates about a fixed axis with variable angular velocity equal to alpha-beta t. The angle through which it rotates

Answer 1

Given, w= alpha-beta*t

at rest , w= alpha - beta *t = 0

=> t = alpha/beta

we know that

angle of rotation = integration(w.dt) , from limit t=0 to t= alpha/beta

                         = integration( (alpha-beta*t).dt

=alpha*t - beta * (t^2) /2,  from limit t=0 to t = alpha/beta

=alpha*(alpha/beta - 0)  - beta * [(alpha/beta)^2 - 0] /2

= (alpha^2)/(2*beta)

Answer 2

Answer:

The angle through which it rotates is $\theta=\dfrac{\alpha^2}{2\beta}$.

Explanation:

Given that,

The angular velocity$\omega=\alpha-\beta t$

At rest, The angular velocity is equal to zero.

$\omega =o$

$\alpha-\beta t=0$

$t=\dfrac{\alpha}{\beta}$

We need to calculate the angle through which it rotates

The angle through which it rotates is

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

$\int_{0}^{\theta}{d\theta}=\int_{0}^{\frac{\alpha}{\beta}}{\omega}dt$

On integration both side

$\theta=\int_{0}^{\frac{\alpha}{\beta}}{(\alpha-\beta t)}dt$

$\theta=[\alpha t-\dfrac{\beta t^2}{2}]_{0}^{\frac{\alpha}{\beta}}$

$\theta=\alpha\times\dfrac{\alpha}{\beta}-\dfrac{\beta(\dfrac{\alpha}{\beta})^2}{2}$

$\theta=\dfrac{\alpha^2}{\beta}-\dfrac{\alpha^2}{2\beta}$

$\theta=\dfrac{\alpha^2}{2\beta}$

Hence, The angle through which it rotates is $\theta=\dfrac{\alpha^2}{2\beta}$.