Question

The rectangular components of acceleration for a particle is ax = 3t and ay= 30-10t where a is in m/s^2. If the particle start from rest from the origin, find the radius of curvature of the path at t=2 sec.

Answer 1

Answer on Question #79789 - Physics - Mechanics, Relativity

The rectangular components of acceleration for a particle is $a_x = 3t$ and $a_y = 30 - 10t$ , where $a$ is in $m/s^2$ . If the particle starts from rest from the origin, find the radius of curvature of the path at $t = 2$ sec.

Solution.

$$\vec {a} = \frac {d \vec {v}}{d t}$$

Initial conditions:

$$v _ {0. x} = 0; v _ {0. y} = 0$$ $$v _ {x} (t) = \int a _ {x} d t = \int 3 t d t = \frac {3}{2} t ^ {2} + v _ {0. x} = \frac {3}{2} t ^ {2}$$ $$v _ {y} (t) = \int a _ {y} d t = \int (3 0 - 1 0 t) d t = 3 0 t - 5 t ^ {2} + v _ {0. y} = 3 0 t - 5 t ^ {2}$$ $$v ^ {2} = v _ {x} ^ {2} + v _ {y} ^ {2}$$ $$v _ {x} (2) = 6 m / s; v _ {y} (2) = 4 0 m / s$$ $$v ^ {2} (2) = 6 ^ {2} + 4 0 ^ {2} = 1 6 3 6 m ^ {2} / s ^ {2}$$

The velocity of the particle is directed along the tangent to its trajectory.

The angle $\theta$ between $\vec{v}$ and the $x$ -axis is:

$$\cos \theta = \frac {v _ {x} (2)}{v (2)} = \frac {6}{\sqrt {1 6 3 6}} = \frac {3}{\sqrt {4 0 9}}; \sin \theta = \frac {v _ {y} (2)}{v (2)} = \frac {4 0}{\sqrt {1 6 3 6}} = \frac {2 0}{\sqrt {4 0 9}}$$

The acceleration vector can be decomposed into $x$ and $y$ components as well as into tangential and normal components (parallel and perpendicular to the tangent to the trajectory of motion). The normal component of the acceleration correlates with the radius of curvature as follows:

$$a _ {n o r m} = \frac {v ^ {2}}{R}$$

It can be expressed in terms of $x$ and $y$ components of the acceleration:

$$a _ {n o r m} = a _ {x} \times \cos (9 0 {}^ {\circ} - \theta) - a _ {y} \times \cos \theta$$ $$a _ {x} (2) = 6 \, \mathrm{m/s^2}; \, a _ {y} (2) = 10 \, \mathrm{m/s^2}$$ $$a _ {norm} (2) = a _ {x} (2) \times \sin \theta - a _ {y} (2) \times \cos \theta = \frac {6 \times 20 - 10 \times 3 \, \mathrm{m}}{\sqrt {409}} \, \mathrm{s^2}$$ $$a _ {norm} (2) = \frac {90}{\sqrt {409}} \, \mathrm{m/s^2}$$

We obtain:

$$R = \frac {v ^ {2} (2)}{a _ {norm} (2)} = \frac {1636 \times \sqrt {409}}{90} = \frac {2 \times 409 ^ {3/2}}{45} \, \mathrm{m} \approx 367.6 \, \mathrm{m}$$

Answer: 367.6 m

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