Question

A disc with its plane horizontal collides with a floor. Before the collision line of velocity vector of the disc makes an angle 60° with normal to the floor and after the collision, velocity vector of the disc becomes perpendicular to its velocity vector before the collision without any change in its magnitude. Find average value of the coefficient of friction between the disc and the floor.

Answer 1

1

Answer 2

To solve this problem, we will use the principles of impulse-momentum and the definition of the coefficient of friction in terms of impulses.

Let the initial velocity of the disc be $\vec{v_1}$ and the final velocity be $\vec{v_2}$. Let the magnitude of the velocity be $v$. According to the problem statement, $|\vec{v_1}| = |\vec{v_2}| = v$. This implies that the kinetic energy of the disc is conserved, meaning the collision is elastic, and the coefficient of restitution, $e$, is 1.

Let's set up a coordinate system. Let the floor be the $xy$-plane, and the normal to the floor be along the $z$-axis.

1. Initial Velocity $\vec{v_1}$: The velocity vector makes an angle of $60^\circ$ with the normal to the floor (z-axis). Let's assume the initial velocity lies in the $xz$-plane for simplicity. This choice does not affect the final result for the coefficient of friction, as shown in the detailed thought process. The vertical component (along $z$-axis) is $v_{1z} = -v \cos(60^\circ) = -v/2$ (negative sign indicates downward motion). The horizontal component (along $x$-axis) is $v_{1x} = v \sin(60^\circ) = v\sqrt{3}/2$. The $y$-component is $v_{1y} = 0$. So, $\vec{v_1} = \left(v\frac{\sqrt{3}}{2}\right)\hat{i} + 0\hat{j} - \left(\frac{v}{2}\right)\hat{k}$.

2. Final Velocity $\vec{v_2}$: We are given two conditions for $\vec{v_2}$: a. $|\vec{v_2}| = v$. b. $\vec{v_2}$ is perpendicular to $\vec{v_1}$, i.e., $\vec{v_1} \cdot \vec{v_2} = 0$.

Let $\vec{v_2} = v_{2x}\hat{i} + v_{2y}\hat{j} + v_{2z}\hat{k}$.

From condition (b): $\left(v\frac{\sqrt{3}}{2}\right)v_{2x} + 0 \cdot v_{2y} - \left(\frac{v}{2}\right)v_{2z} = 0$ Multiplying by $2/v$ (assuming $v \neq 0$): $\sqrt{3}v_{2x} - v_{2z} = 0 \implies v_{2z} = \sqrt{3}v_{2x}$. (Equation 1)

3. Using Coefficient of Restitution: Since the collision is elastic ($e=1$), the relative speed of separation along the normal is equal to the relative speed of approach. $e = -\frac{v_{2z}}{v_{1z}}$ $1 = -\frac{v_{2z}}{(-v/2)} \implies v_{2z} = v/2$.

4. Determining Components of $\vec{v_2}$: Substitute $v_{2z} = v/2$ into Equation 1: $v/2 = \sqrt{3}v_{2x} \implies v_{2x} = \frac{v}{2\sqrt{3}}$.

Now use condition (a), $|\vec{v_2}| = v$: $v_{2x}^2 + v_{2y}^2 + v_{2z}^2 = v^2$ $\left(\frac{v}{2\sqrt{3}}\right)^2 + v_{2y}^2 + \left(\frac{v}{2}\right)^2 = v^2$ $\frac{v^2}{12} + v_{2y}^2 + \frac{v^2}{4} = v^2$ $\frac{v^2}{12} + \frac{3v^2}{12} + v_{2y}^2 = v^2$ $\frac{4v^2}{12} + v_{2y}^2 = v^2$ $\frac{v^2}{3} + v_{2y}^2 = v^2$ $v_{2y}^2 = v^2 - \frac{v^2}{3} = \frac{2v^2}{3}$ $v_{2y} = \pm v\sqrt{\frac{2}{3}}$.

So, the final velocity components are $v_{2x} = \frac{v}{2\sqrt{3}}$, $v_{2y} = \pm v\sqrt{\frac{2}{3}}$, and $v_{2z} = \frac{v}{2}$.

5. Calculating Impulses: The change in momentum is equal to the impulse. The normal impulse $J_N$ acts along the z-axis: $J_N = \Delta p_z = m(v_{2z} - v_{1z}) = m\left(\frac{v}{2} - \left(-\frac{v}{2}\right)\right) = m\left(\frac{v}{2} + \frac{v}{2}\right) = mv$.

The frictional impulse $\vec{J_f}$ acts in the $xy$-plane: $\vec{J_f} = \Delta \vec{p}_{xy} = m(v_{2x} - v_{1x})\hat{i} + m(v_{2y} - v_{1y})\hat{j}$ $J_x = m(v_{2x} - v_{1x}) = m\left(\frac{v}{2\sqrt{3}} - v\frac{\sqrt{3}}{2}\right) = mv\left(\frac{1 - 3}{2\sqrt{3}}\right) = mv\left(\frac{-2}{2\sqrt{3}}\right) = -\frac{mv}{\sqrt{3}}$. $J_y = m(v_{2y} - v_{1y}) = m\left(\pm v\sqrt{\frac{2}{3}} - 0\right) = \pm mv\sqrt{\frac{2}{3}}$.

The magnitude of the frictional impulse is $|\vec{J_f}| = \sqrt{J_x^2 + J_y^2}$: $|\vec{J_f}| = \sqrt{\left(-\frac{mv}{\sqrt{3}}\right)^2 + \left(\pm mv\sqrt{\frac{2}{3}}\right)^2}$ $|\vec{J_f}| = \sqrt{\frac{m^2v^2}{3} + \frac{2m^2v^2}{3}} = \sqrt{\frac{3m^2v^2}{3}} = \sqrt{m^2v^2} = mv$.

6. Calculating the Average Coefficient of Friction: The average coefficient of friction $\mu$ is defined as the ratio of the magnitude of the frictional impulse to the magnitude of the normal impulse: $\mu = \frac{|\vec{J_f}|}{J_N} = \frac{mv}{mv} = 1$.