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Question

A particle is moving in the xy plane with uniform speed v. The equation of trajectory for the particle is represented by the parabola y=ax2, where a is a +ve constant. Find the radius of curvature of trajectory at the point x=0.

A
a2
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B
12a
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C
4a
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D
32a
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Solution

The correct option is B 12a
Point x=0 will represent position at the origin (0,0), since y=0 at x=0 from the equation of trajectory given by y=ax2
Differentiating w.r.t time gives:
dydt=2axdxdt ...(i)
vy=2ax vx ...(ii)

Putting, x=0 in Eq. (i) gives vy=0, hence it will have only velocity in x direction i.e vx at x=0. The motion parameters have been shown in the figure given below:


Particle is moving at uniform speed v
vx=v ...(iii)
Again differentiating Eq. (i) w.r.t time:
d2ydt2=2a[dxdt.dxdt+x.d2xdt2]
ay=2a[(dxdt)2+x.d2xdt2]
At point x=0,
ay=2a[(dxdt)2+0]=2av2x
From Eq. (iii), we get
ay=2av2

As shown in figure at (x=0, y=0), net acceleration of the particle is ay which is to net velocity (vx).
ay=(vx)2R where R is the radius of curvature.
2av2=v2R
R=12a is the radius of curvature of the trajectory at x=0.

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