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So indeed, the torque $\vec \tau = \vec r \times \vec F$ slows this thing down, and assuming that the force is never radial but only tangential, this is just $\tau = r ~ F$ where r is the radius of the circle and $F$ is the magnitude of the braking force. Newton's law $\frac{d\vec p}{dt} = \sum_i \vec F_i$ becomes for angular problems $\frac{d\vec L}{dt} = ... 3 I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ... 0 First Thought (probably not the fastest) Let us assume you have a vector space in$R^{3}$with a quaternion defined as: $$\mathbf{q} = q_{F}^{*} \ q_{B} \\ = a + b \hat{\mathbf{x}} + c \hat{\mathbf{y}} + d \hat{\mathbf{z}}$$ where$(a, b, c, d)$are the Euler parameters and$(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}})\$ defines the reference ...