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

[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?

 

[Satvik Pandey]

Yes, I think so. Good..

Thank you.

Check the sign of C.

I got that.It should be -3g/2.
Please tell me what to do next?

 

[TSny]

Yes, -3g/2.

Think about what you are trying to find and how that is related to y.

 

[ehild]

Thank you.
I got that.It should be -3g/2.
Please tell me what to do next?

The C is not correct yet.

ehild

 

[TSny]

The C is no correct yet.

ehild

Oops. Yes, there's still something missing in the expression for C. Thanks.

 

[Satvik Pandey]

I think it should be -3g/2r.

 

[ehild]

It is all right now. Now remember what y is, and what you need to the solution of the original problem (read it again).

ehild

 

[Satvik Pandey]

I need to find the velocity of block when it makes 90° with horizontal.

So ω²=\frac{9.8}{4}-\frac{29.4}{4}e^{-\pi/2}

What is the value of e^{-\pi/2}.
Sorry but I don't know its value.

Is e^{-\pi/2} equal to 0.021 ?

 

[ehild]

You have got a calculator, don't you? And there is pi on it, and you can calculate the value of an exponential function. Maybe, with inverse ln (shift ln).

And your formula is wrong, r is missing and why is 4 in the denominator of the second term?

ehild

 

[Satvik Pandey]

You have got a calculator, don't you? And there is pi on it, and you can calculate the value of an exponential function. Maybe, with inverse ln (shift ln).

And your formula is wrong, r is missing and why is 4 in the denominator of the second term?

ehild

I substituted r=2.It was given in the question.

Is e^pi/2=11.5703463164?

 

[ehild]

I substituted r=2.It was given in the question.

Is e^pi/2=11.5703463164?

In the question, r was given as 200 cm.

You need e^-pi/2.

 

[Satvik Pandey]

ω²=\frac{9.8}{4}-\frac{29.4}{4}e^{-\pi/2}

I found e^{-\pi/2}=0.0216

On putting values I found I found ω=1.5136

As V=rω

So V=3.02136

S=1.05m

t=\sqrt{\frac{2S}{g}}=0.429

Velocity of block makes 90° with horizontal at the he point at which it leaves the wedge.
So vertical component of V is 0.
So x=Vt=1.40m=140cm.

 

[Satvik Pandey]

In the question, r was given as 200 cm.

You need e^-pi/2.

I changed its unit into meter as the value of 'g' that I have used is 9.8\frac{m}{s²}
So 200cm=2m

 

[ehild]

ω²=\frac{9.8}{4}-\frac{29.4}{4}e^{-\pi/2}

I found e^{-\pi/2}=0.0216

On putting values I found I found ω=1.5136

As V=rω

So V=3.02136

S=1.05m

t=\sqrt{\frac{2S}{g}}=0.429

Velocity of block makes 90° with horizontal at the he point at which it leaves the wedge.
So vertical component of V is 0.
So x=Vt=1.40m=140cm.

It is correct in principle, but the numerical values are not. e^{-\pi/2}=0.20788
How did you get 0.0216? You forgot the parentheses and calculated $e^{-\pi}/2$?

Also, t=0.4629 s instead of 0.429,
and v makes 0° angle with the horizontal, instead of 90° but these are typos.
And you were right about r=2 m, somehow I thought of 200 mm instead of 200 cm.

The problem was far above high-school level, you needed knowledge about differential equations.
You solved it at the end, congratulation, but it would be more useful for you to practice on problems which are more elementary.

ehild

 

[Satvik Pandey]

ω²=\frac{9.8}{4}-\frac{29.4}{4}e^{-\pi/2}

ω=0.96025
As v=rω
So v=1.920

Since t=0.4629
So x=1.920*0.4629
=0.889 m
And this is approximately equal to 89 cm.

Got correct answer

 

[Satvik Pandey]

It is correct in principle, but the numerical values are not. e^{-\pi/2}=0.20788
How did you get 0.0216? You forgot the parentheses and calculated $e^{-\pi}/2$?
ehild

Yes I forgot the parentheses.Sorry for inconvenience.

Also, t=0.4629 s instead of 0.429,
and v makes 0° angle with the horizontal, instead of 90° but these are typos.

ehild

Oh! my typing mistakes. When would they end.

The problem was far above high-school level, you needed knowledge about differential equations.
You solved it at the end, congratulation, but it would be more useful for you to practice on problems which are more elementary.

ehild

How did you find that I am in high-school?.Is this a prediction?
I have solved similar types of question but in those questions friction was not involved.
When I saw this question I developed curiosity.
Yes,this question needed knowledge about differential equations.
And I do not have much knowledge about it.But I have some knowledge about basic integration and differentiation. I found the method to solve linear differential equation in a secondary school mathematics book.I just applied that method and I got correct answer.

it would be more useful for you to practice on problems which are more elementary

I will follow your suggestion.

Thank you ehild for helping me.I couldn't have solved this without your help.

 

[Satvik Pandey]

Thank you ehild,TSny,mfb,CAF123 for helping me to find the solution.

 

[ehild]

How did you find that I am in high-school?.

I saw your Profile

You become a real scientist if you understand what you do. Maths and Physics are not mere collections of recipes. You have to start from the bases and build up your knowledge step by step. Every step must follow logically from your previous knowledge, and then you would feel a new theorem or proof easy as if you could have discovered them by yourself.

ehild

 

[Satvik Pandey]

I saw your Profile

You become a real scientist if you understand what you do. Maths and Physics are not mere collections of recipes. You have to start from the bases and build up your knowledge step by step. Every step must follow logically from your previous knowledge, and then you would feel a new theorem or proof easy as if you could have discovered them by yourself.

ehild

Thank you ehild I will follow your suggestion.