Particle's velocity,acceleration and force

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SUMMARY

The discussion focuses on the analysis of a particle's motion under a force, specifically addressing its velocity, kinetic energy, acceleration, and force. The position vector is defined as x(t) = (r cos(ωt), r sin(ωt), ut). The correct velocity vector is derived as v(t) = (-rωsin(ωt), rωcos(ωt), 1), while the kinetic energy is shown to be constant using the identity sin²x + cos²x = 1. The acceleration vector is a = (-rω²cos(ωt), -r²ωsin(ωt), 0), and the force is calculated as F = m * a, demonstrating that the force's magnitude is constant despite the acceleration not being constant.

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  • Understanding of vector calculus and differentiation
  • Familiarity with Newton's laws of motion
  • Knowledge of kinetic energy and its derivation
  • Basic trigonometric identities, specifically sin²x + cos²x = 1
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Homework Statement



A particle P, with mass m, moves under the influence of a force. At time t the position vector of P, according to observer O in an inertial frame S, is given by x(t)=(r cos(ωt),r sin(ωt),ut)

(i) Differentiate the position vector, to obtain the velocity vector at time t.

(ii) Show that the kinetic energy of P is constant and give its value.

(iii) Differentiate the velocity vector, to obtain the acceleration vector at time t.

(iv) Show that the magnitude of the force acting on P is constant and give its value.

Homework Equations





The Attempt at a Solution



(i) v(t)=(-r ωsin(ωt),r ωcos(ωt),t)

(ii) KE = 1/2* m * v^2

KE is constant as velocity is constant.

v(t)=(-r ωsin(ωt),r ωcos(ωt),t)

v^2(t)= (-r ωsin(ωt),r ωcos(ωt),t) ^2

-r ωsin(ωt), after squaring = -r ^2ω^2 sin^2 (ωt),
r ωcos(ωt),after squaring r^2 ω^2cos^2(ωt),
t after squaring= t^2

KE value= 1/2* m * [-r ^2ω^2 sin^2 (ωt),r^2 ω^2cos^2(ωt),t^2]

(iii) v(t)=(-r ωsin(ωt),r ωcos(ωt),t)

a= [(-r ω^2 cos(ωt),-r^2 ωsin(ωt)]

(iv) F=ma
F= m [(-r ω^2 cos(ωt),-r^2 ωsin(ωt)]

force is constant as accelration is constant.
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are these answers correct?
 
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As you've made a little differentation error in (i) (d(ut)/dt != t), in (ii) you've ended up with a wrong expression for the KE. As you can see, with the t2, your expression is most certainly not constant w.r.t. time.

I should also point out that v2 = |v|2, ie. you should end up with a scalar quantity in (ii). Use the fact that sin2x + cos2x = 1 to prove KE constant.

In (iii) you've dropped out a component of a (it is zero, so this is just a minor point, but it looks as if a is a vector of R2 instead of R3).

As for (iv), you haven't shown that the acceleration is constant (and this would be a difficult task, as it isn't), so you shouldn't really use that to prove the magnitude of the force is constant. Emphasis on "magnitude".
 
Last edited:

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