Question

8. A small sleeve can slide on fixed horizontal ring made of pipe of circular cross section. Radius of the ring is r, mass of the sleeve is m, and coefficient of friction between the sleeve and the ring is $\mu$. The ring is given an initial velocity of $v_0$.

(a) How will the speed of the sleeve vary with time t? (b) How will the speed of the sleeve vary with distance travelled s?

Answer 1

(a) $v(t) = \frac{r v_0}{r + \mu t v_0}$ (b) $v(s) = v_0 e^{-\frac{\mu s}{r}}$

Answer 2

The problem describes a small sleeve sliding on a fixed horizontal circular ring. The sleeve has mass $m$, the ring has radius $r$, and the coefficient of kinetic friction between them is $\mu$. The sleeve is given an initial tangential velocity $v_0$.

1. Analysis of Forces:

• Normal Force (N): As the sleeve moves in a circular path, a centripetal force is required, which is provided by the normal force from the ring acting radially inward on the sleeve.

  $N = \frac{mv^2}{r}$

• Friction Force ($f_k$): This force opposes the tangential motion of the sleeve and acts tangentially.

  $f_k = \mu N = \mu \frac{mv^2}{r}$

2. Equation of Motion in Tangential Direction:

The friction force causes the sleeve to decelerate. Applying Newton's second law in the tangential direction:

$F_{tangential} = ma_{tangential}$

$-f_k = m \frac{dv}{dt}$

Substituting the expression for $f_k$:

$-\mu \frac{mv^2}{r} = m \frac{dv}{dt}$

$\frac{dv}{dt} = -\frac{\mu v^2}{r}$

(a) Speed as a function of time t:

We have the differential equation: $\frac{dv}{dt} = -\frac{\mu v^2}{r}$

Separate variables and integrate:

$\frac{dv}{v^2} = -\frac{\mu}{r} dt$

Integrate from initial velocity $v_0$ at $t=0$ to $v$ at time $t$:

$\int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} -\frac{\mu}{r} dt$

$\left[-\frac{1}{v}\right]_{v_0}^{v} = -\frac{\mu}{r} [t]_{0}^{t}$

$-\frac{1}{v} - \left(-\frac{1}{v_0}\right) = -\frac{\mu t}{r}$

$\frac{1}{v_0} - \frac{1}{v} = -\frac{\mu t}{r}$

$\frac{1}{v} = \frac{1}{v_0} + \frac{\mu t}{r}$

$\frac{1}{v} = \frac{r + \mu t v_0}{r v_0}$

$v(t) = \frac{r v_0}{r + \mu t v_0}$

(b) Speed as a function of distance travelled s:

We can relate tangential acceleration to distance travelled using $a_{tangential} = v \frac{dv}{ds}$.

From the equation of motion: $m v \frac{dv}{ds} = -f_k$

$m v \frac{dv}{ds} = -\mu \frac{mv^2}{r}$

Assuming $v \neq 0$, we can divide by $mv$:

$\frac{dv}{ds} = -\frac{\mu v}{r}$

Separate variables and integrate:

$\frac{dv}{v} = -\frac{\mu}{r} ds$

Integrate from initial velocity $v_0$ at $s=0$ to $v$ at distance $s$:

$\int_{v_0}^{v} \frac{dv}{v} = \int_{0}^{s} -\frac{\mu}{r} ds$

$[\ln|v|]_{v_0}^{v} = -\frac{\mu}{r} [s]_{0}^{s}$

$\ln v - \ln v_0 = -\frac{\mu s}{r}$

$\ln\left(\frac{v}{v_0}\right) = -\frac{\mu s}{r}$

$\frac{v}{v_0} = e^{-\frac{\mu s}{r}}$

$v(s) = v_0 e^{-\frac{\mu s}{r}}$