Question

A ring of mass M and radius R is kept on a smooth horizontal surface. An insect of mass M begins to move along the circumference of the ring with a speed v relative to the ring. The velocity of center of the ring after a time t is

• A. $\frac{vcos(vt/R)}{2}$
• B. $\frac{vesin(vt/R)}{2}$
• C. $\frac{vesin(vt/2R)}{2}$
• D. zero

Answer 1

Answer: (B)

2

Answer 2

The problem involves a system of a ring and an insect. Since the surface is smooth (no friction) and no external horizontal forces act on the system, the total horizontal momentum of the system must be conserved.

1. Conservation of Momentum:

Let $M$ be the mass of the ring and $R$ its radius. The mass of the insect is also $M$.

Initially, both the ring and the insect are at rest. Therefore, the initial total momentum of the system is zero.

$P_{initial} = M_{ring} \vec{V}_{ring, initial} + M_{insect} \vec{V}_{insect, initial} = M(0) + M(0) = 0$.

Let $\vec{V}_{ring}$ be the velocity of the center of the ring at time $t$.

Let $\vec{v}_{I/R}$ be the velocity of the insect relative to the ring. Its magnitude is given as $v$.

The velocity of the insect relative to the ground, $\vec{V}_{insect, ground}$, is the vector sum of the ring's velocity and the insect's velocity relative to the ring:

$\vec{V}_{insect, ground} = \vec{V}_{ring} + \vec{v}_{I/R}$.

According to the conservation of momentum:

$P_{final} = M\vec{V}_{ring} + M\vec{V}_{insect, ground} = 0$

Substitute $\vec{V}_{insect, ground}$:

$M\vec{V}_{ring} + M(\vec{V}_{ring} + \vec{v}_{I/R}) = 0$

$2M\vec{V}_{ring} + M\vec{v}_{I/R} = 0$

$\vec{V}_{ring} = -\frac{1}{2}\vec{v}_{I/R}$

This equation shows that the velocity of the center of the ring is always half the magnitude of the insect's velocity relative to the ring, and in the opposite direction. The magnitude of the velocity of the center of the ring is therefore $|\vec{V}_{ring}| = \frac{1}{2}|\vec{v}_{I/R}| = \frac{v}{2}$.

However, none of the options are a constant $v/2$. This implies the question is asking for a specific component of the velocity, which depends on the choice of coordinate system and the insect's starting position and direction of motion.

2. Determining the Components of $\vec{v}_{I/R}$:

Let's assume a standard coordinate system where the center of the ring is initially at the origin $(0,0)$.

Let the insect start at the point $(R,0)$ on the circumference and move counter-clockwise.

In time $t$, the insect moves an angular displacement $\theta = \omega t = (v/R)t$ relative to the center of the ring.

The position vector of the insect relative to the ring's center at time $t$ is $\vec{r}_{I/R}(t) = (R\cos\theta, R\sin\theta)$.

The velocity of the insect relative to the ring, $\vec{v}_{I/R}$, is the time derivative of its position vector:

$\vec{v}_{I/R} = \frac{d}{dt}(R\cos\theta \hat{i} + R\sin\theta \hat{j})$

Since $\theta = vt/R$, we have $\frac{d\theta}{dt} = \frac{v}{R}$.

$\vec{v}_{I/R} = (-R\sin\theta \frac{d\theta}{dt})\hat{i} + (R\cos\theta \frac{d\theta}{dt})\hat{j}$

$\vec{v}_{I/R} = (-R\sin(\frac{vt}{R}) \frac{v}{R})\hat{i} + (R\cos(\frac{vt}{R}) \frac{v}{R})\hat{j}$

$\vec{v}_{I/R} = -v\sin(\frac{vt}{R})\hat{i} + v\cos(\frac{vt}{R})\hat{j}$

3. Velocity of the Center of the Ring:

Now substitute this expression for $\vec{v}_{I/R}$ into the momentum conservation equation $\vec{V}_{ring} = -\frac{1}{2}\vec{v}_{I/R}$:

$\vec{V}_{ring} = -\frac{1}{2}(-v\sin(\frac{vt}{R})\hat{i} + v\cos(\frac{vt}{R})\hat{j})$

$\vec{V}_{ring} = \frac{v}{2}\sin(\frac{vt}{R})\hat{i} - \frac{v}{2}\cos(\frac{vt}{R})\hat{j}$

The components of the velocity of the center of the ring are:

$V_{ring,x} = \frac{v}{2}\sin(\frac{vt}{R})$

$V_{ring,y} = -\frac{v}{2}\cos(\frac{vt}{R})$

Comparing these with the given options, option (2) matches the x-component of the velocity of the center of the ring. If the question implies asking for a specific component (e.g., the x-component), then option (2) is the correct answer under this common interpretation of initial conditions.