# Block slipping against wedge

1. Aug 17, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data

Consider a fixed wedge, whose part is a quarter circle of radius as shown. A small block of mass 'm' slides down the wedge and leaves horizontally from a height 'h' as shown.The coefficient of friction between wedge and block is 'μ$_{k}$' . Compute the value of in centimetres , where 'x' is the distance from bottom of wedge of the point where the block falls given the values below:
Here is the figure.
https://d18l82el6cdm1i.cloudfront.net/solvable/9f0d7815b5.e8a0649017.E9Nd6g.png
_Details : _

R=200cm
h=105cm
μ$_{k}$=0.5
m=2g

2. Relevant equations

3. The attempt at a solution

I think we have to apply law of conservation of energy here.
i.e., potential energy of block
(E$_{p}$) =work done by friction (W$_{f}$) + Kinetic energy (E$_{k}$)....(1)
Here, I have problem in finding the W$_{f}$.
In know F$_{f}$ = μR.
Here, the normal reactionary force which is being exerted on the block on the wedge is not constant. I tried few ways but I am unable to find W$_{f}$. It would be nice if somebody could help.

Last edited: Aug 17, 2014
2. Aug 17, 2014

### Staff: Mentor

And it depends on the velocity. If the acceleration depends on the velocity, a good idea is to set up a differential equation and try to solve that.

3. Aug 17, 2014

### Satvik Pandey

Thanks for the quick reply mfb.

I tried this--

dw=μrdθ(mgsinθ+mrω$^{2}$)

I know that ω=dθ/dt.

But putting this does not help.
Is there any other way through which we can find ω?
This is the figure.

Last edited: Aug 17, 2014
4. Aug 17, 2014

### Staff: Mentor

What is w?

Can you express everything in terms of θ, ω and dω/dt?

5. Aug 17, 2014

### Satvik Pandey

dw infinitely small work done by frictional force.

6. Aug 17, 2014

### Satvik Pandey

I think I can.

As a=αr and α=dω/dt
gcosθ-gsinθ-rω^2=rdω/dt

Is it right?

7. Aug 17, 2014

### Staff: Mentor

Work is not relevant here.

Somehow the μ got lost, apart from that it looks good.

8. Aug 17, 2014

### Satvik Pandey

gcosθ-μgsinθ-μrω^2=rdω/dt.

What to do next?

9. Aug 17, 2014

### Satvik Pandey

Is eq.1 in my post#1 correct?

10. Aug 17, 2014

### Staff: Mentor

Do some mathematical magic to try to find a solution.
As a homework problem, it should have a reasonable solution, I just don't see it now.

11. Aug 17, 2014

### CAF123

Hi Satvik,
Yes, it is. I think you forgot the contribution of the work done by gravity along the circular arc though in #3.

I seem to be having the same problem as you. I can write down an explicit expression for the work done by all the external forces but there is one term in it that I cannot integrate.

12. Aug 17, 2014

### Satvik Pandey

Hi CAF123
Work done by gravity is mgrsinθ.Is it right?

13. Aug 17, 2014

### CAF123

I think I should write out the problem I am having.

The work done on the mass is due to friction and gravity. An infinitesimal displacement of the mass along the arc is $\text{d} \underline s = R\theta \hat s$ using the definition of $\theta$ in your attachment. The arc is now parametrised in terms of $\theta$ and so $$W = \int_0^{\pi/2} \underline{F} \cdot \text{d} \underline s = \underbrace{-\int_0^{\pi/2} \mu\left(mg \sin \theta + \frac{mv(\theta)^2}{R}\right)R\text{d}\theta}_{\text{work done by friction}}+ \underbrace{\int_0^{\pi/2} mg \cos \theta R \text{d}\theta}_{\text{work done by gravity}}$$ This can be rewritten like $$W = -R\mu mg \int_0^{\pi/2} \sin \theta \text{d} \theta - mR^2 \int_0^{\pi/2} \dot{\theta}^2 \text{d}\theta + mgR\int_0^{\pi/2} \cos \theta \text{d}\theta$$

It is this middle integral that I cannot evaluate. The last integral is mgR, which we could have obtained simply by noting gravity is a conservative force and the mass falls a vertical distance R. The 'work done by friction' brace is the finite version of your dw I believe.

14. Aug 17, 2014

### TSny

Try the trick of using the chain rule to write dω/dt = (dω/dθ)(dθ/dt)

15. Aug 17, 2014

### CAF123

Hi TSny,
Do you have any suggestions on how to compute the middle integral in my last post? Should I find $\dot{\theta}$ first?

16. Aug 17, 2014

### TSny

I don't see a way to do the integral. It seems to me that solving the differential equation in Satvik's post #8 is the way to go.

If you like work and energy, you can get an equivalent differential equation by considering the differential form of the work-energy theorem: dW = d(mv2/2) and using your expressions for the work done by friction and gravity for an infinitesimal displacement ds = Rdθ. This way, you avoid getting a differential equation with a time derivative. (In Satvik's equation, you have a time derivative which can be rewritten in terms of a derivative with respect to θ using the chain rule.)

θ is the independent variable of the diff. eq.
For the dependent variable, y say, choose either y = v2 or y = ω2.

Last edited: Aug 17, 2014
17. Aug 17, 2014

### CAF123

Yes, so the equation becomes $$\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta$$ So, I could solve this for $\dot{\theta}(\theta)$ and input into my W-E energy equation.

Last edited: Aug 17, 2014
18. Aug 17, 2014

### TSny

Yes. But, of course, if you can solve the differential equation $\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta$ for $\omega(\theta)$, you essentially have the answer for the speed at the bottom of the track. There is no need to go back to the W-E equation.

19. Aug 17, 2014

### CAF123

Yes, of course, thanks. Could you elaborate on what you meant by the below argument?
I could take the differential of the equation W, and I think then the first and third terms of my equation would vanish (they are constants) but I am not sure what the middle term would look like. Or did you mean something else?
Thanks!

20. Aug 17, 2014

### haruspex

Fwiw, maybe not much, I simulated it. Assuming the exit speed is proportional to sqrt(rg), plotting for mu up to 0.55 gives an excellent fit to $v = \sqrt{rg} 1.9187(0.75 - \mu)$. No idea what the magic constant relates to. Presumably the 0.75 indicates the threshold at which it sticks somewhere on the wedge.
Goes a bit wrong for higher mu, but that might be rounding errors.