Logo
Logo

Anisotropic friction

A1  0.50 At what angle $\alpha_1$ to the $X$ axis should the body velocity vector be for the absolute value of the power of the friction force be at maximum?

__
Power of the friction force is given by the equation:
$$P=\left(F,v\right)=-\left(\mu_xmg\cos^2\alpha+\mu_ymg\sin^2\alpha\right)v$$
since $\mu_x>\mu_y$, then maximum of the power is reached when $\alpha=0$
Ответ: $$\alpha=0$$
A2  0.50 At what angle $\alpha_2$ to the $X$ axis should the body velocity vector be for the absolute value of the power of the friction force be 1.2 times less than maximum?

__
From the previous point it follows that:
$$\mu_xmgv\cos^2\alpha+\mu_ymgv\sin^2\alpha=\frac{1}{1.2} \mu_xmgv$$
From where we get:
$$
\cos \alpha_2 = \pm \sqrt{\frac{5\mu_x - 6\mu_y}{6(\mu_x - \mu_y)}} = \pm \frac{1}{\sqrt{2}} \\
\sin \alpha_2 = \pm \sqrt{\frac{\mu_x}{6(\mu_x-\mu_y)}} = \pm \frac{1}{\sqrt{2}}
$$
Ответ: $$\alpha_2=\pm \pi/4; \pm 3\pi/4$$
A3  1.00 Let the initial velocity have components $v_{0x} = 1 \text{m/s}$ and $v_{0y} = 1 \text{m/s}$. After some time the velocity component along the $Y$ axis equals $v_{1y} = 0,\!25 \text{m/s}$. What is the velocity magnitude at this moment?

__
Newton's equation:
\begin{cases}
m\dfrac{dv_x}{dt} &=-\mu_xmg\dfrac{v_x}{v}\\
m\dfrac{dv_y}{dt} &=-\mu_ymg\dfrac{v_y}{v}
\end{cases}
Dividing one equation by another, we get:
$$\dfrac{dv_x}{dv_y}=\dfrac{\mu_xv_x}{\mu_yv_y}$$
Integrating these equations, we obtain:
$$\dfrac{v_x^{\mu_y}}{v_y^{\mu_x}}=Const$$
From the initial conditions we find:
$$v_x=0.125 \text{m/s}$$
So the absolute value of velosity is
Ответ: $$v=0.28 \text{m/s}$$
A4  1.00 Let the velocity be $v_2 = 1.0 \text{m/s}$. At what angle $\alpha_3$ to the $X$ axis should the velocity vector be for the radius of curvature of the trajectory be minimum? What is this radius equal to? The free fall acceleration is $g=9,\!8 \text{m/s}^2$.

__
Projection of the friction force onto the direction perpendicular to the velocity is given by equation:
$$F_n=mg(\mu_x-\mu_y)\sin\alpha\cos\alpha$$
Newton's second law projected onto the direction perpendicular to the velocity has the form:
$$\dfrac{mv^2}{R}=mg(\mu_x-\mu_y)\sin\alpha\cos\alpha,$$
where $R$ is the radius of curvature of the trajectory. It's obvious that
$$R=\dfrac{v^2}{g(\mu_x-\mu_y)\sin\alpha\cos\alpha}=\dfrac{2v^2}{g(\mu_x-\mu_y)\sin2\alpha}$$
The minimum radius of curvature will be when:
Ответ: $$\alpha=\pi/4$$
and
$$R_{min}=\dfrac{2v^2}{g(\mu_x-\mu_y)}$$
A5  1.00 In a single diagram on the $XY$ plane, sketch the trajectories of the body launched at the angles $\alpha_4 = \pi/6$ and $\alpha_5 = \pi/3$ for the friction coefficients specified above. The magnitudes of initial velocities are the same. Solve the same problem for the friction coefficients $ \mu_x = 0,\! 4$ and $\mu_y = 0,\!7$.

__
If $\mu_x=\mu_y$, then the accelerations along the x and y axes will be proportional to the ratio of the initial velocities and the body will move in a straight line. If $\mu_x>\mu_y$, then the speed along the x-axis will decrease faster than in the previous case and the body will deviate from a linear motion as shown in the figure. The direction of deflection does not depend on how the initial velocity is directed. If $\mu_x<\mu_y$, then the body will deviate in the opposite direction.
Ответ:
B1  2.00 A body of mass $m$ is at rest at the origin. A force has been applied to it at an angle $\alpha$ to the $X$ axis. The force magnitude $F(t)=\gamma t$ linearly grows with time. Find the dependence of the moment the body starts moving on $\alpha$. Ignore the stagnation phenomenon.

__
Let us notice that the body will start the motion with the angle $\varphi \neq \alpha$. Then projections of the forces at this moment are given by

$$\begin{cases}F_{\text{fr},x} = \mu_x mg \cos \varphi = \gamma t\cos \alpha \\
F_{\text{fr},y} = \mu_y mg \sin \varphi = \gamma t\sin \alpha \end{cases}$$

Dividing these equations by $\mu_x mg$ and $\mu_y mg$ correspondingly and summing up squares, we get

$$(mg)^2 = (\gamma t)^2 \left[ \left(\frac{\cos \alpha}{\mu_x} \right)^2 + \left(\frac{\sin \alpha}{\mu_y} \right)^2 \right]$$

So the answer is
Ответ: $$t=\dfrac{\mu_x\mu_ymg}{\gamma\sqrt{\mu_y^2\cos^2\alpha+\mu_x^2\sin^2\alpha}}$$
C1  1.50 For a given initial velocity $ v_0 $ find the dependence of its velocity $ v $ on the angle of rotation of the rod $ \varphi $ assuming that the other body remains at rest.

__
Newton's law projected onto the direction of motion of a point-like mass is:
$$m\dfrac{dv}{dt}=-mg(\mu_x\cos^2\varphi+\mu_y\sin^2\varphi)$$
Since $v=L\varphi$, then:
$$L\ddot\varphi=-g(\mu_x\cos^2\varphi+\mu_y\sin^2\varphi)$$
Multiply this equation by $\dot\varphi\, dt$ we get
$$L\dot\varphi d\dot\varphi=-g(\mu_x\cos^2\varphi+\mu_y\sin^2\varphi)d\varphi$$
Integrating this equation, we obtain:
Ответ: $$v^2+gL\left((\mu_x+\mu_y)\varphi+(\mu_x-\mu_y)\dfrac{\sin2\varphi}{2}\right)=v_0^2$$
C2  1.50 Find the maximum value of the initial velocity $ v_{0\mathrm{max}}$ at which the other body will remain at rest.

__
Newton's law for a moving body in projection onto a rod has the form:
$$T+F_{\text{fr}}\sin\beta=m\dot\varphi^2L$$
where $\beta$ is the angle between frictional force and direction of velocity.
Using the result from C1, we find that:
$$T=\frac{mv_0^2}{L}-mg(\mu_x+\mu_y)\varphi-mg(\mu_x-\mu_y)\sin2\varphi$$
Using the result from B1, we find that:
$$\frac{mv_0^2}{L}-mg(\mu_x+\mu_y)\varphi-mg(\mu_x-\mu_y)\sin2\varphi\le\dfrac{\mu_x\mu_y}{\sqrt{\mu_y^2\sin^2\varphi+\mu_x^2\cos^2\varphi}}mg$$
Note that with increasing $\varphi$, the left hand side decreases, and the right hand side increases, so the body starts to move right at $\varphi=0$. So
Ответ: $$v_\text{0max}=\sqrt{\mu_ygL}=2.2 \text{m/s}$$
C3  1.00 What distance will the body travel until it stops completely if the initial velocity is $ v_{0\mathrm{max}} $?

__
Substituting $v_\text{0max}$ from the previous section in the final equation of the C1, we get the equation for the $\varphi$ at the stopping point $(v=0)$:

$$gL\left((\mu_x+\mu_y)\varphi+(\mu_x-\mu_y)\dfrac{\sin2\varphi}{2}\right)=\mu_y gL $$

Numerically solving this equation, we get the answer for the $\varphi$, and so the travelled distance is equal to
Ответ: $$L \varphi = 0.34 \text{m}$$