Motion of a ParticleDate: 3/31/96 at 3:26:4 From: Anonymous Subject: differential equation I'd like to know how to solve a problem in which a particle is shot from the ground at an initial known angle and velocity. The force of friction is f=av^3 where {a} is a constant. The only other force is the force of gravity. Thank you for any assistance you can provide. Date: 4/3/96 at 11:31:44 From: Doctor Sebastien Subject: Re: differential equation Hi, Start doing the problem by studying the motion of the particle along the horizontal and along the vertical separately. Let's suppose that the particle is shot at a velocity V and at an angle of x to the horizontal. The initial horizontal speed is given by V cos x, and the initial vertical speed is given by V sin x. These two terms are calculated from the fact that sine of an angle is opposite side/ hypotenuse and cosine of an angle is adjacent side/hypotenuse. Let v be the velocity of the particle at time t. Now, let's study horizontal motion. The force, f1, on the object is given by f1 = - av^3. The force is negative since it is in the opposite direction to motion. From Newton's second law, force = mass * acceleration. Therefore, acceleration = force/mass = f/m, where m is mass. Acceleration of object is rate of change of velocity with respect to time. Acceleration = dv/dt Therefore dv/dt = - (a/m) v^3 The above equation describes motion of the particle. From this you can find an equation giving velocity at time t. Now let's consider the vertical motion. When the particle is moving upwards, the force on it = -(weight + friction) = - (mg + av^3) When the particle is moving downwards, the force on it = weight - friction = mg - av^3 Friction is always against motion. I have not solved the problem completely. You can solve the problem given the acceleration in terms of velocity. Use variable separables and integrate with respect to t. -Doctor Sebastien, The Math Forum |
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