Dynamics Lecture: Kinematics using Polar Coordinates
Dynamics Lecture: Kinematics using Polar Coordinates

Get 5 free video unlocks on our app with code GOMOBILE

Snapsolve any problem by taking a picture.
Try it in the Numerade app?

Use the angular velocity to derive the expression for the acceleration of a particle in polar coordinates in Eq. (3.33).
rÎ¸^2 + (2rÎ¸ + rÎ¸)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 eÎ¸ 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

This problem has been solved!

Try Numerade free for 7 days

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 Î¸(t) is the angular coordinate. The parameter t represents time. Knowing that r = r(t) and Î¸ = Î¸(t), where r is the unit vector in the radial direction and Î¸ 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)Î¸'(t),a = (r”(t) – r(t)Î¸'(t)^2)r(t) + (2r'(t)Î¸'(t) + r(t)Î¸”(t))Î¸(t).

Notation: The prime denotes derivative with respect to time. Remark: v is the tangential velocity (v = r'(t)), Ï‰ is the angular velocity (Ï‰ = Î¸'(t)), ac is the centripetal acceleration (ac = r(t)Î¸'(t)^2), and cor is the Coriolis acceleration (cor = 2r'(t)Î¸'(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 $b \omega^{2}$.

Oops! There was an issue generating an instant solution

Enter your parent or guardian’s email address:

Already have an account? Log in

Create an account to get free access

or

EMAIL

PASSWORD

You are watching: SOLVED: Use the angular velocity to derive the expression for the acceleration of a particle in polar coordinates in Eq. (3.33). rÎ¸^2 + (2rÎ¸ + rÎ¸)e (3.33) Show that the product rule for vector diff. Info created by Bút Chì Xanh selection and synthesis along with other related topics.