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Kinematics of a Particle in a Plane (using intrinsic description)

Example: At time t_o the velocity vector and acceleration vector of a particle, P, are as shown in the figure below. The speed of P is v_P = 2 ft/sec. The magnitude of the acceleration of P is

a_P = 40 ft/sec², and θ = 30^o. Find the tangential component of acceleration, the normal component

of acceleration, and the radius of curvature of the path.

Solution:

The velocity of P is v_P = v_P e_t = 2 e_t ft/sec since the velocity is always tangent to its path.

Here e_t is the unit tangential vector. The acceleration vector, a_P, will have a normal and tangential component. Consider the figure below where e_n is the unit normal vector and points toward the

center of curvature.

From the figure a_Pt = a_p cosθ = 40 cos 30 = 34.6 ft/sec² in the direction shown.

and a_Pn = a_p sinθ = 40 sin 30 = 20 ft/sec² in the direction shown.

Now a_Pn = v²/ρ so ρ = v²/ a_Pn = 2² / 20 = 1/5 ft

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