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Use the angular velocity to derive the expression for the acceleration of a particle in polar coordinates in Eq. (3.33).
rθ^2 + (2rθ + rθ)e
(3.33)
Show that the product rule for vector differentiation (see Table 3.1) holds for the vector cross product.
Derive the vector derivative of eθ in Eq. (3.30) Id (3.30) 1p
Show that the radius of curvature of a path defined by the function y(x) in Cartesian coordinates is given by Eq. (3.50).
1 + R
(3.50)
dx
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01:21
Polar Coordinates: The trajectory of a particle is specified in polar coordinates as a function of time as V = r + 0 = log(1 + V^2).1.1 Find formulas for the velocity and acceleration of the particle as a function of V and t using the polar coordinate system. Find a formula for the speed of the particle in terms of V.1.2 Hence, find the normal and tangential components of acceleration of the particle.1.3 Hence, find a formula for the radius of curvature of the path as a function of t (or r).
04:49
In polar coordinates, the position of a point particle is described by the vector r(t). Where r(t) is the radial coordinate and θ(t) is the angular coordinate. The parameter t represents time. Knowing that r = r(t) and θ = θ(t), where r is the unit vector in the radial direction and θ is the unit vector in the angular direction, use the chain rule to show that the velocity and acceleration vectors are, respectively:
v = r'(t) + r(t)θ'(t),a = (r”(t) – r(t)θ'(t)^2)r(t) + (2r'(t)θ'(t) + r(t)θ”(t))θ(t).
Notation: The prime denotes derivative with respect to time. Remark: v is the tangential velocity (v = r'(t)), ω is the angular velocity (ω = θ'(t)), ac is the centripetal acceleration (ac = r(t)θ'(t)^2), and cor is the Coriolis acceleration (cor = 2r'(t)θ'(t)).
01:20
Consider a particle \end{tabular} moving on a circular path of radius $b$ described by $\mathbf{r}(t)=b \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j}$ where $\omega=d \theta / d t$ is the constant angular velocity.Find the velocity vector and show that it is orthogonal to $\mathbf{r}(t)$.
01:53
Consider a particle \end{tabular} moving on a circular path of radius $b$ described by $\mathbf{r}(t)=b \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j}$ where $\omega=d \theta / d t$ is the constant angular velocity.Show that the magnitude of the acceleration vector is $b \omega^{2}$.
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