Finding Work Done by Friction on a Sliding Block Against a Wedge

[Satvik Pandey]

Homework Statement

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

Homework Equations

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.

 

[mfb]

Here, the normal reactionary force which is being exerted on the block on the wedge is not constant.

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.

 

[Satvik Pandey]

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.

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.

 

[mfb]

What is w?

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

 

[Satvik Pandey]

What is w?

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

dw infinitely small work done by frictional force.

 

[Satvik Pandey]

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

I think I can.

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

Is it right?

 

[mfb]

Work is not relevant here.

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

 

[Satvik Pandey]

Work is not relevant here.

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

gcosθ-μgsinθ-μrω^2=rdω/dt.

What to do next?

 

[Satvik Pandey]

Work is not relevant here.

Is eq.1 in my post#1 correct?

 

[mfb]

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.

 

[CAF123]

Hi Satvik,

Is eq.1 in my post#1 correct?

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.

 

[Satvik Pandey]

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.

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

 

[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.

 

[TSny]

gcosθ-μgsinθ-μrω^2=rdω/dt.

What to do next?

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

 

[CAF123]

Hi TSny,

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

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

 

[TSny]

Hi TSny,

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

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.

 

[CAF123]

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.)

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.

 

[TSny]

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.

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.

 

[CAF123]

Yes, of course, thanks. Could you elaborate on what you meant by the below argument?

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.

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!

 

[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.

 

[mfb]

It has to scale with sqrt(rg) for dimensional reasons. Setting r=g=1, WolframAlpha can solve the differential equation.

Setting $\theta = \frac{\pi}{2}$, I get

$$v = \sqrt{2} \frac{\sqrt{-e^{-\mu \pi} (2 \mu^2 e^{\mu \pi} - e^{\mu \pi} +3 \mu)}}{\sqrt{4 \mu^2+1}}$$

This is well-defined up to μ of about 0.603. This value corresponds to a stopped block exactly at the lower edge.

Testing this with μ=0 gives v=sqrt(2), in agreement with the expectation.

Setting μ=0 in the formula of haruspex gives v=1.439, not so far away but not the correct value.

 

[TSny]

Assuming the exit speed is proportional to sqrt(rg), plotting for mu up to 0.55 gives an excellent fit to $v = \sqrt{rg} 1.9187(0.75 - \mu)$.

Unless I'm overlooking something, the differential equation can be solved. I get an exit speed that is proportional to √(rg), but the coefficient is a more complicated function of μ. I find that the block doesn't make it to the bottom if μ is larger than about 0.603.

[oops! I didn't see mfb's post of the solution above. Glad to see we get the same result.]

 

[TSny]

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!

Replace the upper limits of integration by an arbitrary angle θ. Then take the derivative with respect to this angle θ using the second fundamental theorem of calculus

 

[TSny]

If you let y = v², then you get a linear, first-order differential equation in y that can be solved using an integrating factor.

 

[Satvik Pandey]

As
$$\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta$$

So

ω$^{2}$=$\frac{gcosθdθ-μgsinθdθ-rωdω}{μmrdθ}$


Putting this value in in post#2 I got

dw=μmgsinθdθ+rgcosθdθ-μrgsinθdθ-r$^{2}$ωdω

I can integrate both side to find W.But it contains a expression involving ω.
How to get rid of that?

 

[TSny]

As
$$\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta$$

So

ω$^{2}$=$\frac{gcosθdθ-μgsinθdθ-rωdω}{μmrdθ}$


Putting this value in in post#2 I got

dw=μmgsinθdθ+rgcosθdθ-μrgsinθdθ-r$^{2}$ωdω

I can integrate both side to find W.But it contains a expression involving ω.
How to get rid of that?

The terms in your second and third equations above are not dimensionally consistent.

Go back to the first equation: $$\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta$$
In the first term, can you write $\omega \frac{\text{d}\omega}{\text{d}\theta}$ as the derivative of some quantity? That is, $$\omega \frac{\text{d}\omega}{\text{d}\theta} = \frac{d (?)}{d\theta}$$

 

[ehild]

As
$$\omega r \frac{\text{d}\omega}{\text{d}\theta} + \mu r \omega^2 = g \cos \theta - \mu g \sin \theta$$

Notice that

Spoiler

$ω\frac{\text{d}\omega}{\text{d}\theta} =0.5 \frac{\text{d}(\omega^2)}{\text{d}\theta}$

As it was suggested by TSny use ω²=y.

Then you have a first order linear differential equation for y.

$$0.5 r\frac{\text{d}y}{\text{d}\theta} + \mu r y= g \cos \theta - \mu g \sin \theta$$

Solve for y, using the initial condition y=0 at θ=0.

If you are unfamiliar with differential equations yet, try the solution in the form $$y=Ce^{kθ}+A\cos(θ)+B\sin(θ)$$

Find the constants k, A, B which make the equation valid for any θ. Then Fit C to the initial condition.

ehild

Edit:
Sorry TSny, I did not see your post when I wrote my one.

 

[Satvik Pandey]

Notice that

Spoiler

$ω\frac{\text{d}\omega}{\text{d}\theta} =0.5 \frac{\text{d}(\omega^2)}{\text{d}\theta}$

As it was suggested by TSny use ω²=y.

Then you have a first order linear differential equation for y.

$$0.5 r\frac{\text{d}y}{\text{d}\theta} + \mu r y= g \cos \theta - \mu g \sin \theta$$

Solve for y, using the initial condition y=0 at θ=0.

If you are unfamiliar with differential equations yet, try the solution in the form $$y=Ce^{kθ}+A\cos(θ)+B\sin(θ)$$

Find the constants k, A, B which make the equation valid for any θ. Then Fit C to the initial condition.

ehild

Edit:
Sorry TSny, I did not see your post when I wrote my one.

I have not studied differential equations yet but I have made an attempt to solve it.
AS
$$0.5 r\frac{\text{d}y}{\text{d}\theta} + \mu r y= g \cos \theta - \mu g \sin \theta$$

I think it is a linear differential equation.

Here integrating factor is

e$^{\intμ/0.5 dθ}$ = e$^{θ}$ (As μ=0.5)

ye$^{θ}$=$\int e^{θ}.\frac{gcosθ-μgsinθ}{0.5r}$ + C

I can integrate RHS with the help of integration by parts but I am not sure if I am right till here.

Is it right till here,ehild?

 

[Satvik Pandey]

Notice that

Spoiler

$ω\frac{\text{d}\omega}{\text{d}\theta} =0.5 \frac{\text{d}(\omega^2)}{\text{d}\theta}$

As it was suggested by TSny use ω²=y.

As ω² = y
dω²/dω =dy/dω
2ω = dy/dω
dω = dy/2ω
Putting this value on ωr$\frac{dω}{dθ}$
I got r$\frac{dy}{2dθ}$.

 

[ehild]

I looks OK, (but substitute μ=0.5 in front of the sin term, too) but you can choose the other method for the solution of a linear differential equation. The solution is equal to Y=Yh+ Yp where Yh is the general solution of the homogeneous part 0.5ry'+μry=0 and Yp is a particular solution of the original equation. In case the right hand side is a sum of sins and cosines, you can try a particular solution also as a linear combination of sins and cosines. It is a bit more elementary than the other method, involving integrals.

ehild

 

[ehild]

As ω² = y
dω²/dω =dy/dω
2ω = dy/dω
dω = dy/2ω
Putting this value on ωr$\frac{dω}{dθ}$
I got r$\frac{dy}{2dθ}$.

Yes, it is correct, and remember it. You will find this relation useful for a great many cases.

ehild

 

[Satvik Pandey]

I looks OK, (but substitute μ=0.5 in front of the sin term, too) but you can choose the other method for the solution of a linear differential equation. The solution is equal to Y=Yh+ Yp where Yh is the general solution of the homogeneous part 0.5ry'+μry=0 and Yp is a particular solution of the original equation. In case the right hand side is a sum of sins and cosines, you can try a particular solution also as a linear combination of sins and cosines. It is a bit more elementary than the other method, involving integrals.

ehild

ye$^{θ}$=$\int e^{θ}.\frac{gcosθ-μgsinθ}{0.5r}$ + C

$\frac{g}{0.5r}$$\int e^{θ}cosθdθ$-0.5$\int e^{θ}sinθdθ$

Through integration by parts I found

$\int e^{θ}cosθdθ$=e^θ(sinθ +cosθ)/2

and $\int e^{θ}sinθdθ$=e^θ(sinθ -cosθ)/2Putting this value I got

=$\frac{ge^{θ}}{2r}(sinθ+3cosθ)$+C

ye$^{θ}$=$\frac{ge^{θ}}{2r}(sinθ+3cosθ)$+C

Is it right till here?

On putting θ=0 I found C=3g/2.

 

[TSny]

ye$^{θ}$=$\frac{ge^{θ}}{2r}(sinθ+3cosθ)$+C

Is it right till here?

Yes, I think so. Good.

On putting θ=0 I found C=3g/2.

Check the sign of C.

 

[ehild]

ye$^{θ}$=$\frac{ge^{θ}}{2r}(sinθ+3cosθ)$+C

Is it right till here?

On putting θ=0 I found C=3g/2.

Almost. You need y, so the soultion of the differential equation is $y=\frac{g}{2r}(sinθ+3cosθ)+Ce^{-θ}$
With the initial condition y(0)=0, $C=\frac{-3g}{2r}$

You did not try my method

ehild

 

[Satvik Pandey]

Almost. You need y, so the soultion of the differential equation is $y=\frac{g}{2r}(sinθ+3cosθ)+Ce^{-θ}$
With the initial condition y(0)=0, $C=\frac{-3g}{2r}$ehild

At initial point y(ω²) and θ are equal to zero.So C=-3g/2

You did not try my method

ehild

Sorry.

ehild and TSny, could you please tell me what to do next?