Question

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

a. Find an expression for $ay$, the vertical component of acceleration, in terms of $x$, $vx$, and $ax$. Hint: Use the basic definitions of velocity and acceleration.

b. Suppose the bead is released at some negative value of$x$ and has a speed of $2.3m/s$ as it passes through the lowest point of the parabola. What is the net force on the bead at this instant? Write your answer in component form

Short answer

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

b). The component form this force is$\overrightarrow{F}=(2.12\mathrm{N})\hat{j}$.

Step-by-step solution

Given Infromation (Part a)

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

Explanation (Part a) 

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,

$V_y=4x\times V_x$

Again differentiating on both sides:

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

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

Hence, the expression for$a_y=4m^{-1}(v_x^2+x{a}_x)$

Final Answer  (part a)

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

Given Infromation (Part b)

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

Explanation (Part b)

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,

$a_y=4a_x\times x+4\times v_x^2$

Substituting above values:

$a_y=4\left(0\right)\times a_x+4(2.3{)}^2$

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

Calculate the net force on the bead

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" $=0.1\times 21.16$

$=2.116\mathrm{N}$

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

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 $\overrightarrow{F}=\left(2.12\mathrm{N}\right)\widehat{j}$

Final Answer

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