A particle moves, so that its position vector is given by r $$=$$ $$cos$$ωt $$x^$$ $$+$$$$sin$$ωt $$y^$$ , where ω is a constant. Which of the following is true?

Question

A particle moves, so that its position vector is given by r $$=$$ $$cos$$ωt $$x^$$ $$+$$$$sin$$ωt $$y^$$ , where ω is a constant. Which of the following is true?

Infinity Learn · Accepted Answer

Position vector of the particle is given byr $$=$$ $$cos$$ωtx⏞ $$+$$$$sin$$ωty⏞ where, ω is a constant.Velocity of the particle $$isv=drdt=ddt($$$$cos$$ω$$tx^+$$$$sin$$ω$$ty^)$$   $$=(-$$$$sin$$ω$$t)$$ω$$x^+($$$$cos$$ω$$t)$$ω$$y^$$   $$=-$$ω$$($$$$sin$$ω$$tx^-$$$$cos$$ω$$ty^)Acceleration$$ of the particle,$$a=dvdt=ddt(-$$ω$$sin$$ω$$tx^+$$ω$$cos$$ω$$ty^)=-$$ω$$2cos$$ω$$tx^-$$ω$$2sin$$ω$$ty^$$⇒  $$a=-$$ω$$2r=$$ω$$2(-r)Assuming$$ the particle is at P, then its position vector is directed as shown in the diagram.Therefore, acceleration is directed towards $$-r$$, i.e. towards O (origin).v·$$r=-$$ω$$($$$$sin$$ω$$tx^-$$$$cos$$ω$$ty^)$$·$$($$$$cos$$ω$$tx^+$$$$sin$$ω$$ty^)=-$$ω[$$sin$$ωt·$$cos$$ω$$t+0+0-$$$$sin$$ωt·$$cos$$ωt]$$=-$$ω$$(0)=0$$⇒ v ⊥ rThus, velocity is perpendicular to r.

A particle moves, so that its position vector is given by $r = cosω t \hat{x} + sinω t \hat{y}$ , where $ω$ is a constant. Which of the following is true?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

A particle moves, so that its position vector is given by r = cosωt x^ +sinωt y^ , where ω is a constant. Which of the following is true?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

A particle moves, so that its position vector is given by $r = cosω t \hat{x} + sinω t \hat{y}$ , where $ω$ is a constant. Which of the following is true?