Question

A particle of mass $m$ moves in a circular orbit with $x = R\cos(\omega t)$ and $y = R\sin(\omega t)$ observed in inertial frame $S_1$. Another frame $S_2$ moves with velocity $\vec{v} = \omega R \hat{i}$ with respect to $S_1$, and origins coincide at $t=0$. The angular momentum at $t = \frac{2\pi}{\omega}$ as observed in $S_2$ about its origin is $(mR^2\omega)x$. Then $x$ is ............. (Specify answer up to two digits after decimal.)

Hint:

When changing frames by Galilean transformation, subtract the frame velocity before computing cross products.

Answer 1

Correct Answer: 5.25

Answer should be : 5.25-5.30 or -5.30--5.25
Step 1: Find velocity in $S_2$.
In $S_1$, velocity is $\vec{v}_1 = (-R\omega\sin\omega t,\, R\omega\cos\omega t)$.
Frame $S_2$ moves with velocity $(\omega R, 0)$, so
$\vec{v}_2 = \vec{v}_1 - (\omega R, 0)$.

Step 2: Evaluate at one full period.
At $t = 2\pi/\omega$, position becomes $(R, 0)$, same as at $t=0$.
Velocity in $S_1$ returns to $(0, R\omega)$.
Thus, in $S_2$: $\vec{v}_2 = (-\omega R,\, R\omega)$.

Step 3: Compute angular momentum.
$\vec{L} = \vec{r} \times m\vec{v}_2$.
$\vec{r} = (R,0)$, $m\vec{v}_2 = m(-\omega R,\, \omega R)$.
Cross product magnitude: $L = mR(\omega R) = mR^2\omega$.

Step 4: Compare with given form.
Given $L = (mR^2\omega)x$, so $x = 1.00$.