Question

Obtain and expression relating torque with angular acceleration for a rigid body.

Answer 1

$\tau = I\alpha$

Answer 2

The expression relating torque with angular acceleration for a rigid body is derived as follows:

Consider a rigid body rotating about a fixed axis with angular acceleration $\alpha$. Imagine the body is composed of many small particles.

1. Force on a mass element: Consider a small mass element $dm$ located at a perpendicular distance $r$ from the axis of rotation. According to Newton's second law, the tangential force $dF_t$ acting on this element is: $dF_t = dm \cdot a_t$ where $a_t$ is the tangential acceleration of the mass element.

2. Tangential and angular acceleration relation: The tangential acceleration $a_t$ is related to the angular acceleration $\alpha$ by: $a_t = r\alpha$

3. Substitute $a_t$: Substituting the expression for $a_t$ into the force equation: $dF_t = dm \cdot (r\alpha)$

4. Torque due to the element: The torque $d\tau$ produced by this tangential force about the axis of rotation is the product of the force and its perpendicular distance from the axis: $d\tau = r \cdot dF_t$ Substituting the expression for $dF_t$: $d\tau = r \cdot (dm \cdot r\alpha)$ $d\tau = r^2 \alpha \cdot dm$

5. Total torque: To find the total torque $\tau$ acting on the entire rigid body, we integrate the torques due to all such mass elements over the entire mass of the body: $\tau = \int d\tau = \int r^2 \alpha \cdot dm$ Since $\alpha$ is constant for all particles in a rigid body rotating about a fixed axis, it can be taken out of the integral: $\tau = \alpha \int r^2 dm$

6. Moment of Inertia: The term $\int r^2 dm$ is defined as the moment of inertia ($I$) of the rigid body about the given axis of rotation. $I = \int r^2 dm$

7. Final Expression: Substituting $I$ into the equation for torque, we obtain the desired expression: $\tau = I\alpha$

This expression is the rotational analogue of Newton's second law ($F = ma$), where torque ($\tau$) is the analogue of force ($F$), moment of inertia ($I$) is the analogue of mass ($m$), and angular acceleration ($\alpha$) is the analogue of linear acceleration ($a$).

Explanation of the solution:

For a rigid body rotating about a fixed axis:

1. A small mass element $dm$ at distance $r$ from the axis experiences a tangential force $dF_t = dm \cdot a_t$.
2. The tangential acceleration $a_t = r\alpha$, where $\alpha$ is the angular acceleration.
3. So, $dF_t = dm \cdot r\alpha$.
4. The torque $d\tau$ due to this element is $d\tau = r \cdot dF_t = r \cdot (dm \cdot r\alpha) = r^2 \alpha \cdot dm$.
5. Integrating over the entire body, the total torque $\tau = \int r^2 \alpha \cdot dm$.
6. Since $\alpha$ is constant, $\tau = \alpha \int r^2 dm$.
7. Defining the moment of inertia $I = \int r^2 dm$, we get the relation: $\tau = I\alpha$