Question

\({ }^{\dagger}\) A rigid body has angular velocity \(\boldsymbol{\omega}\) and has one point \(O\) fixed relative to a frame \(\tilde{R}\). Show that if \(\boldsymbol{\omega} \wedge \tilde{D} \boldsymbol{\omega} \neq 0\), then \(O\) is the only point with zero acceleration relative to \(\tilde{R}\).

Short answer

Answer: If the condition \(\boldsymbol{\omega} \wedge \tilde{D} \boldsymbol{\omega} \neq 0\) holds, the only point on the rigid body with zero acceleration relative to the frame \(\tilde{R}\) is the fixed point \(O\).

Step-by-step solution

Express the velocity of a point in the rigid body

We'll first express the velocity \(\boldsymbol{v}\) of a point \(P\) on the rigid body relative to the fixed point \(O\). Since the body is rigid, we know that the distance between any two points in the body doesn't change over time. So, considering the vector \(\boldsymbol{r}\) from point \(O\) to point \(P\), we have: $$ \boldsymbol{v} = \frac{d \boldsymbol{r}}{dt} $$ Now, we note that we can express the derivative of \(\boldsymbol{r}\) as the angular velocity \(\boldsymbol{\omega}\) crossed with the vector \(\boldsymbol{r}\): $$ \boldsymbol{v} = \boldsymbol{\omega} \wedge \boldsymbol{r} $$

Express the acceleration of a point in the rigid body

Next, we'll find the acceleration \(\boldsymbol{a}\) of the point \(P\) on the rigid body. To do this, we will differentiate the velocity \(\boldsymbol{v}\) of the point \(P\) with respect to time: $$ \boldsymbol{a} = \frac{d \boldsymbol{v}}{dt} $$ So substituting the expression for \(\boldsymbol{v}\) from step 1, we get: $$ \boldsymbol{a} = \frac{d}{dt} (\boldsymbol{\omega} \wedge \boldsymbol{r}) $$ Using the product rule for differentiation, we have: $$ \boldsymbol{a} = \tilde{D} \boldsymbol{\omega} \wedge \boldsymbol{r} + \boldsymbol{\omega} \wedge \tilde{D} \boldsymbol{r} $$ Since \(\tilde{D} \boldsymbol{r} = \boldsymbol{v}\), this becomes: $$ \boldsymbol{a} = \tilde{D} \boldsymbol{\omega} \wedge \boldsymbol{r} + \boldsymbol{\omega} \wedge \boldsymbol{v} $$

Determine the condition for zero acceleration

Now, we want to find the condition under which the acceleration \(\boldsymbol{a}\) is zero: $$ \boldsymbol{0} = \tilde{D} \boldsymbol{\omega} \wedge \boldsymbol{r} + \boldsymbol{\omega} \wedge \boldsymbol{v} $$ The expression for zero acceleration can be rewritten as: $$ \boldsymbol{v} = - \frac{\tilde{D} \boldsymbol{\omega} \wedge \boldsymbol{r}}{\boldsymbol{\omega}} $$ Now, if \(\boldsymbol{\omega} \wedge \tilde{D} \boldsymbol{\omega} \neq 0\), then this implies that the only point on the rigid body with zero acceleration is the point \(O\), which is fixed relative to the frame \(\tilde{R}\). Thus, if \(\boldsymbol{\omega} \wedge \tilde{D} \boldsymbol{\omega} \neq 0\), the only point on the rigid body with zero acceleration relative to \(\tilde{R}\) is the fixed point \(O\).

Key concepts

Angular Velocity

Angular velocity, denoted by the symbol \( \boldsymbol{\omega} \) is a vector quantity that represents the rate of rotation of a rigid body around a fixed axis. It's not just how fast something spins; it's also in which direction. Imagine a spinning wheel: its angular velocity tells us how rapidly it rotates and which way it is spinning.

The direction of the angular velocity vector follows the right-hand rule: if you curl the fingers of your right hand in the direction of the rotation, your thumb points along the angular velocity vector. In the context of the given problem, we can see that \( \boldsymbol{\omega} \) is crucial to determining the motion and acceleration of every point in the rigid body. As we increase or decrease \( \boldsymbol{\omega} \) or change its direction, we affect the velocities of points in the body in a predictable manner. It's the starting point for analyzing many kinds of rotational motion.

Rigid Body Acceleration

Acceleration in the context of a rigid body extends the concept of acceleration as we encounter it with particles. While for particles we simply look at the change in velocity over time, for a rigid body we also need to consider the body's rotational characteristics. For any point in a rigid body, its acceleration \( \boldsymbol{a} \) can have two components: one due to the change in the angular velocity \( \tilde{D} \boldsymbol{\omega} \) and another due to the rotation itself.

To find the total acceleration of any point in the rigid body, we add the acceleration due to the body's angular acceleration and the acceleration due to the body's rotation. In the problem's solution steps, we see that the acceleration of a point \( P \) is given by the derivative of its velocity, which involves a cross product between the angular velocity and the radial vector \( \boldsymbol{r} \) from the fixed point to the point in question. This cross-product encapsulates how the rotation imposes motion on points in the body.

Vector Product in Physics

The vector product, also known as the cross product, is a mathematical operation that takes two vectors and produces a third vector that is perpendicular to the plane of the original vectors. In physics, vector products are essential for describing rotational motion and the effects of forces on rotation.

The cross product's magnitude is proportional to the sine of the angle between the two vectors, implying that if the vectors are parallel, their cross product is zero. When solving for rigid body motion, the cross product of angular velocity \( \boldsymbol{\omega} \) and a radial vector \( \boldsymbol{r} \) yields a new vector representing the velocity of a point on the body. Similarly, the product rule used in the solution steps involves cross products that represent how one vector changes in the direction of another. Vector products are fundamentally important in understanding the interplay between rotational speeds, directions, and the resulting forces within a system.