Question

If $x=Rsin\omega t+R\omega t$ and $y=Rcos\left(\omega t\right)+R$ (where $\omega$ and R are constants), what are x and y components of acceleration when y is minimum?

• A. $0,R{\omega }^2$
• B. $R{\omega }^2,0$
• C. $0,-R{\omega }^2$
• D. $-R{\omega }^2,0$

Answer 1

The correct option is A

$0,R{\omega }^2$

$y=Rcos\omega t+R$

For y to be minimum,

$\frac{dy}{dt}=-\omega Rsin\omega t=0$

$sin\omega t=0$

or $\omega t=0,\pi ,2\pi ,\dots$

$\frac{d^{2}y}{d{T}^2}{\omega }^{2}Rcos\omega t$

At $\omega t=\pi$

$\frac{d^{2}y}{d{t}^2}=+{\omega }^{2}R>0$

Hence y is minimum at $\omega t=\pi$

$a_x=-{\omega }^{2}Rsin\pi =0$

$a_y={\omega }^{2}Rcos\pi ={\omega }^{2}R.$