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A particle moves so that its position vector is given by r = x cos(ωt) î + y sin(ωt) ĵ, where ω is a constant. Which of the following is true?
hard
Motion in a Plane
2016
physics
Velocity is perpendicular to r and acceleration is directed towards the origin.
Velocity is perpendicular to r and acceleration is directed from the origin.
Velocity and acceleration both are perpendicular to r
Velocity and acceleration both are parallel to r.
Explanation
To solve this problem, we need to analyze the motion of the particle given by the position vector r=xcos(ωt)i^+ysin(ωt)j^.
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Sign up / Login First, let's find the velocity vector
The velocity
is the time derivative of the position vector
r.v=dtdr=dtd(xcos(ωt)i^+ysin(ωt)j^) Using the chain rule, we get:
v=−xωsin(ωt)i^+yωcos(ωt)j^ Next, let's find the acceleration vector
The acceleration
is the time derivative of the velocity vector
v.a=dtdv=dtd(−xωsin(ωt)i^+yωcos(ωt)j^) Again, using the chain rule, we get:
a=−xω2cos(ωt)i^−yω2sin(ωt)j^ Now, let's analyze the direction of
and
1. Velocity
is given by:
v=−xωsin(ωt)i^+yωcos(ωt)j^ The dot product of
and
is:
v⋅r=(−xωsin(ωt)i^+yωcos(ωt)j^)⋅(xcos(ωt)i^+ysin(ωt)j^)=−x2ωsin(ωt)cos(ωt)+y2ωcos(ωt)sin(ωt)=ωsin(ωt)cos(ωt)(−x2+y2) Since
sin(ωt)cos(ωt)=0 and
x2=y2, the dot product is zero, indicating that
is perpendicular to
2. Acceleration
is given by:
a=−xω2cos(ωt)i^−yω2sin(ωt)j^ This can be rewritten as:
a=−ω2(xcos(ωt)i^+ysin(ωt)j^)=−ω2r This shows that the acceleration
is directed towards the origin, as it is opposite to
Therefore, the correct option is:
Option 1: Velocity is perpendicular to
and acceleration is directed towards the origin.
However, the provided correct option is 2, which is incorrect based on the analysis.