91Ó°ÊÓ

A $100g$bead slides along a frictionless wire with the parabolic shape $y=\left(2{\mathrm{m}}^{-1}\right)x^2$.

According to the information, the equation for parabolic shape is :

$y=\left(2m^{−1}\right)x^2$

Differentiating on both sides:

$\frac{dy}{dt}=2\left(2x\right)\frac{dx}{dt}$

Consider

$\frac{dy}{dt}=V_y,\frac{dx}{dt}=V_x$

Where, $V_y$is the velocity in $y$direction

$V_x$ is the velocity in$x$direction

Therefore,

$V_y=4x×V_x$

Again differentiating on both sides:

$\frac{d{v}_y}{dt}=4x×\frac{d{v}_x}{dt}+4×\frac{dx}{dt}{v}_x$

Consider

$\frac{d{v}_y}{dt}=a_y,\frac{d{v}_x}{dt}=a_x$

Where,

$a_y$is the acceleration in $y$direction

$a_x$is the acceleration in$x$direction

An expression for$a_y=4{\mathrm{m}}^{-1}\left(v_x^2+x{a}_x\right)$.

A $100g$ bead slides along a frictionless wire with the parabolic shape $y=\left(2{\mathrm{m}}^{-1}\right)x^2$.

According to the information,

$m$is the mass of the bead slides$=0.1kg$

role="math" localid="1648350378763" $V_x$is the velocity in x direction $=2.3$

$V_y$is the velocity in $y$direction$=0$

Here,

$a_y=4a_x×x+4×v_x^2$

Substituting above values:

$a_y=4\left(0\right)×a_x+4(2.3{)}^2$

$=21.16\mathrm{m}/{\mathrm{s}}^2$

Let's find the net force on the bead

$F=m{a}_y$

where,$F=$net force

$m=$mass

role="math" localid="1648350883060" $a_y=$acceleration in the $y$axis

Therefore,

role="math" localid="1648350944319" $=0.1×21.16$

$=2.116\mathrm{N}$

$x=\sqrt{\frac{y}{2}}$

Differentiating on both sides with $t$

$\frac{du}{dt}=\frac{1}{2\sqrt{2}}\frac{vy}{\sqrt{y}}$

$a_x=\frac{1}{2\sqrt{2}}\left(\frac{2y{a}_x−v_y^2}{2y\sqrt{y}}\right)$

$a_t=(0,0)$

$a_x=0$

$F_{netx}=0$

Therefore,

$F_{\text{net}x},F_{\text{nety}}=(0,2.116\mathrm{N})$

The component form of the force is $\stackrel{→}{F}=\left(2.12\mathrm{N}\right)\widehat{j}$

The component form of the force is$\stackrel{→}{F}=\left(2.12\mathrm{N}\right)\widehat{j}$