Question

A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is µ. The block is given an initial speed $v_0$. As a function of the speed v write (a) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration $\left(\frac{dv}{dt}=v\frac{dv}{ds}\right)$ to obtain the speed of the block after one revolution.

Answer 1

(a) Normal force: $N = \frac{mv^2}{R}$ (b) Frictional force: $f = \frac{\mu mv^2}{R}$ (c) Tangential acceleration: $a_t = -\frac{\mu v^2}{R}$ (d) Speed after one revolution: $v_f = v_0 e^{-2\pi \mu}

Answer 2

(a) The normal force $N$ exerted by the wall on the block provides the necessary centripetal force for the horizontal circular motion. Thus, $N = \frac{mv^2}{R}$. (b) The frictional force $f$ is kinetic friction, given by $f = \mu N$. Substituting the expression for $N$, we get $f = \mu \left(\frac{mv^2}{R}\right) = \frac{\mu mv^2}{R}$. This force acts tangentially, opposing the motion. (c) The tangential acceleration $a_t$ is caused by the net force in the tangential direction. According to Newton's second law, $m a_t = -f$. Therefore, $m a_t = -\frac{\mu mv^2}{R}$, which gives $a_t = -\frac{\mu v^2}{R}$. (d) Using the relation $a_t = v \frac{dv}{ds}$, we have $v \frac{dv}{ds} = -\frac{\mu v^2}{R}$. Separating variables yields $\frac{dv}{v} = -\frac{\mu}{R} ds$. Integrating from initial speed $v_0$ at $s=0$ to final speed $v_f$ at $s=2\pi R$ (one revolution), we get $\int_{v_0}^{v_f} \frac{dv}{v} = \int_{0}^{2\pi R} -\frac{\mu}{R} ds$. This simplifies to $\ln\left(\frac{v_f}{v_0}\right) = -2\pi \mu$, so $v_f = v_0 e^{-2\pi \mu}$.