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This is a problem from the problem book by IE Irodov (Problem 1.106). We have a rough inclined plane with coefficient of friction $\mu = \tan\alpha$. A particle is kept on the incline and projected with an initial velocity $v_0$ 'sideways' to the incline, that is into the the plane of the figure shown below.

A coordinate system is set up in the problem with the x-axis pointing down the incline and the y-axis along the initial velocity vector (the origin is at the initial position of the particle). We need to figure out the magnitude of the velocity as a function of the angle $\phi$ its vector makes with the x-axis.

enter image description here

I did the problem in the following manner:

We may fix the velocity vector at the origin and resolve components of the forces acting on the particle as shown:

enter image description here

$v$ is the velocity at an arbitrary time, $F_2$ is the frictional force ($\mu mg\cos\alpha=mg\sin\alpha$), and $F_1$ is the component of gravity along the incline ($mg\sin\alpha$). Note that $\angle DAB$ is $\phi$. Now, we can figure out the tangential acceleration,

$$-\frac{dv}{dt} = g\sin\alpha(1-\cos\phi)\tag1$$

and the radial acceleration

$$v\frac{d\phi}{dt}=-g\sin\alpha\sin\phi\tag2$$

We can introduce a change of variables in $(1)$

$$\frac{dv}{dt} = \frac{dv}{d\phi}\cdot\frac{d\phi}{dt} = \frac{-g\sin\alpha\sin\phi}{v}\cdot\frac{dv}{d\phi}$$

we can substitute this into $(1)$ to get,

$$\frac{dv}{v} = (\csc\phi-\cot\phi)d\phi$$

Integrating yields a solution of the form,

$$\boxed{v=v_0e^{f(\phi)}}$$

where $f$ is a function of $\phi$

But, when I saw the solution, the problem was solved quite simply: we can see that the acceleration along x-axis is $g\sin\alpha(1-\cos\phi)$, which is equal in magnitude to the tangential acceleration. Thus, $v$ and $v_x$ will at any point differ by the same constant value $C$

$$v-v_x = C$$

This $C$ can be easily determined from the initial conditions. This yields,

$$v = \frac{v_0}{1+\cos\phi}$$

Where did I go wrong?

EDIT: This is another figure I'm including just to clarify the situation presented in the problem. Here are different views of the incline for an arbitrary $\phi$.

enter image description here

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  • $\begingroup$ Something seems odd in your question. You say the initial velocity is perpendicular to the incline, i.e., at an angle of $\alpha$ counterclockwise from vertical. Are you sure? $\endgroup$
    – Bill N
    Mar 27, 2015 at 3:55
  • $\begingroup$ @BillN: No, I'm afraid you've misunderstood. The particle is projected in the plane of the incline, it is not projected into the air. By perpendicular to the incline I meant, its perpendicular to the inclined direction. See the first figure I've included. $\vec{v_0}$ is into the plane of the figure with the tail at the large black dot. Perhaps, I should say, the particle is projected sideways onto the plane of the incline. I've edited the question to make it clearer. $\endgroup$
    – Gerard
    Mar 27, 2015 at 4:05
  • $\begingroup$ Oh, you mean into the page. The new figures help quite a bit. $\endgroup$
    – Bill N
    Mar 27, 2015 at 4:09

1 Answer 1

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I don't think you went wrong anywhere, you just didn't find the solution as quickly as the book. Using an integral table, you see that $$ \int\cot\phi= |\ln(\sin\phi)|+C \quad,\quad\int\csc\phi= |\ln(\csc\phi-\cot\phi)|+C.$$ So, integrating $\frac{dv}{v}=(\csc\phi-\cot\phi)d\phi$ and ignoring the absolute values gives $$ v=Ce^{\ln (\csc\phi-\cot\phi)-\ln(\sin\phi)}=C\frac{\csc\phi-\cot\phi}{\sin\phi} $$ $$= C\frac{1-\cos\phi}{\sin^2\phi}= C\frac{1-\cos^2\phi}{(1+\cos\phi)\sin^2\phi}=C\frac{1}{1+\cos\phi}.$$ Your solution is equivalent to the one in the book, you just need some simplification steps which weren't quite obvious.

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