Question

A small block of mass m is placed in circular thali of radius r affixed on a horizontal tabletop. The block is brought close to rim of the thali and given a velocity $v_0$ in tangential direction of the rim. Friction coefficient between the vertical rim of the thali and the block is $\mu$ but the bottom of the thali is frictionless. Determine the following.

(a) How will the speed of the block vary with time t?

(b) How will the speed of the block 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 block experiences a normal force from the rim, which provides the centripetal force ($N = mv^2/r$). This normal force results in a kinetic friction force ($f_k = \mu N$) acting tangentially, opposing the motion. This friction force causes the block to decelerate.

(a) By applying Newton's second law in the tangential direction ($m \frac{dv}{dt} = -f_k$) and substituting $f_k = \mu (mv^2/r)$, we get a differential equation $\frac{dv}{dt} = -\frac{\mu v^2}{r}$. Integrating this equation with initial condition $v(0) = v_0$ yields $v(t) = \frac{r v_0}{r + \mu t v_0}$.

(b) Similarly, using the tangential acceleration as $v \frac{dv}{ds}$ in the equation of motion ($m v \frac{dv}{ds} = -f_k$), we obtain $\frac{dv}{ds} = -\frac{\mu v}{r}$. Integrating this equation with initial condition $v(0) = v_0$ gives $v(s) = v_0 e^{-\frac{\mu s}{r}}$.