A rigid body rotates about a fixed axis with variable angular velocity \( \omega = \alpha - \beta t \) at time \( t \), where \( \alpha, \beta \) are constants. The angle through which it rotates before it stops is:
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For angular motion, the equation:
\[
\omega^2 = \omega_0^2 + 2\beta \theta
\]
is analogous to linear kinematics and helps in solving rotation problems efficiently.
Step 1: {Determine stopping condition}
The angular velocity is given as:
\[
\omega = \alpha - \beta t
\]
The body stops when \( \omega = 0 \), so:
\[
0 = \alpha - \beta t
\]
Solving for \( t \):
\[
t = \frac{\alpha}{\beta}
\]
Step 2: {Use kinematic equation for angular motion}
The equation:
\[
\omega^2 = \omega_0^2 + 2\beta \theta
\]
Substituting \( \omega_0 = \alpha \) and \( \omega = 0 \):
\[
0 = \alpha^2 - 2\beta \theta
\]
Step 3: {Solve for \( \theta \)}
\[
\theta = \frac{\alpha^2}{2\beta}
\]
Thus, the correct answer is (A) \( \frac{\alpha^2}{2\beta} \).
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Approach Solution -2
Given:
The angular velocity of a rigid body varies with time as:
\[
\omega(t) = \alpha - \beta t
\]
Where \( \alpha, \beta \) are positive constants.
Objective:
Find the total angular displacement (angle rotated) before the body comes to rest.
Step 1: Find the time when the body stops rotating
The body stops when angular velocity becomes zero:
\[
\omega = \alpha - \beta t = 0 \Rightarrow t = \frac{\alpha}{\beta}
\]
Step 2: Use integration to find angular displacement
Angular displacement \( \theta \) is given by:
\[
\theta = \int_0^{t} \omega(t)\, dt = \int_0^{\alpha/\beta} (\alpha - \beta t)\, dt
\]
Now integrate:
\[
\theta = \left[ \alpha t - \frac{\beta t^2}{2} \right]_0^{\alpha/\beta}
= \alpha \cdot \frac{\alpha}{\beta} - \frac{\beta}{2} \cdot \left( \frac{\alpha^2}{\beta^2} \right)
= \frac{\alpha^2}{\beta} - \frac{\alpha^2}{2\beta}
= \frac{\alpha^2}{2\beta}
\]
Final Answer:
The angle through which the body rotates before it stops is:
\[
\boxed{\frac{\alpha^2}{2\beta}}
\]