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**Question:**
The video shows a wheelchair moving on a floor. Three points a, b, c are shown at the rim of the three wheels. Wheels with points a and b are rigidly fixed with each other with the second wheel having a somewhat smaller radius than the first. The speeds of a, b, c at some instant are $V_a$, $V_b$ and $V_c$. The angular speeds of the three wheels are $\omega_a$, $\omega_b$, $\omega_c$. The radii of the wheels are $R_a$, $R_b$ and $R_c$.

Choose the correct option(s)

- $V_a = V_b$
- $V_a < V_b$
- $V_a > V_b$
- $V_a/V_b = R_a/R_b$
- $V_a/V_b = R_b/R_a$

- $V_a = V_c$
- $V_a < V_c$
- $V_a > V_c$
- $V_a/V_c = R_a/R_c$
- $V_a/V_c = R_b/R_c$

- $\omega_a = \omega_b$
- $\omega_a < \omega_b$
- $\omega_a > \omega_b$
- $\omega_a/\omega_b = R_a/R_b$
- $\omega_a/\omega_b = R_b/R_a$

- $\omega_a = \omega_c$
- $\omega_a < \omega_c$
- $\omega_a > \omega_c$
- $\omega_a/\omega_c = R_a/R_c$
- $\omega_a/\omega_c = R_c/R_a$

**Analysis:** Since wheels with points a and b are rigidly fixed, we get
\begin{align}
\omega_a=\omega_b.\nonumber
\end{align}
The wheels are rolling without slipping. The velocities of the wheel's contact point with the ground are zero. There is no relative motion between the centres of the wheels (as centres are fixed with the frame). Thus, the velocities of the centres of the wheels are equal. Equating the velocities of the centre of the wheels having points a and c gives,
\begin{align}
\omega_c R_c = \omega_a R_a.\nonumber
\end{align}
Also, $R_a > R_b > R_c$.

**Solution 1:** The question is silent about the reference used for the speeds $V_a$, $V_b$, and $V_c$. The options given in questions (1) and (2) make sense if these speeds are the speeds observed by the man (not relative to the ground). The man and the wheelchair's frame are moving at the same speeds.

In this frame, centre of the wheels are at rest. The wheels are undergoing pure rotation. The speeds of the point a and b are given by
\begin{align}
V_a & =\omega_a R_a \nonumber\\
& = \omega_b R_a \quad\text{(as $\omega_a = \omega_b $)} \nonumber\\
& > \omega_b R_b \quad\text{(as $R_a > R_b $)} \nonumber\\
& = V_b \nonumber
\end{align}
Thus, $V_a > V_b$ and $V_a / V_b = R_a /R_b$. Options (C) and (D) are correct.

The situation will be more complex if the speeds $V_a$, $V_b$, and $V_c$ are defined relative to the ground. The wheel's contact point with the ground (say p) is instantaneously at rest. Let $r_a$ and $r_b$ be the distances of the points a and b at some instant $t$. The speeds of points a and b at this instant are given by
\begin{align}
V_a & = \omega_a r_a, \nonumber\\
V_b & = \omega_b r_b. \nonumber
\end{align}
Note that $\omega_a = \omega_b$, $r_a$ varies from $0$ to $2R_a$ and $r_b$ varies from $R_a-R_b$ to $R_a + R_b$. The values of $V_a$ and $V_b$ will vary with time. The relationship between these two will not be as simple as given in the options.

Let q be the common centre of the wheels with points a and b and point a touches the ground at $t = 0$. The line qb makes an angle $\theta_0$ with qa at $t=0$. The wheels rotate with the same angular velocities $\omega_a = \omega_b = \omega $. At time $t$, the line qa makes an angle $\omega t$ with the vertical and the line qb makes an angle $\omega t + \theta_0$ with the vertical. The speeds of the point a and b at this instant are given by
\begin{align}
V_a & = \omega r_a = 2\omega R_a \sin (\omega t/2), \nonumber\\
V_b & = \omega r_b = 2\omega \sqrt{R_a^2 + R_b^2 - 2 R_a R_b \cos(\omega t + \theta_0)}. \nonumber
\end{align}
We encourage you to plot $V_a$ and $V_b$ for some suitable values of $R_a$, $R_b$ and $\theta_0$.

**Solution 2:** Relative to the man, the centres of the wheel are at the rest and the contact points with the ground move at the same speed. Thus, $V_a = V_c$ i.e., option (A) is correct.

If the speeds are defined w.r.t. the ground then the relationship between $V_a$ and $V_c$ will not be as simple as that given in the options.

**Solution 3:** $\omega_a = \omega_b$. Thus, option (A) is correct.

**Solution 4:** $\omega_a < \omega_c$ and $\omega_a/\omega_c = R_c/R_a$. Thus, option (B) and (E) are correct.

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