Mechanics Doubt HELP!

Here are some problems.if you have solved any one or more please post it's solutions.
The question number 20,28,3120, 28,31 are resolved.
If anyone want to see my attempt for a particular problem, they can ask me I will happily show the attempt.
Thanks in advance.

Note by Talulah Riley
3 weeks, 1 day ago

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Is the answer to problem 27 1.5mg1.5 mg?

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath YES\Huge YES
I am even not able to understand what the question wants to say?
Can you help me to understand what it is saying?

Talulah Riley - 2 weeks, 6 days ago

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I will post a note later.

Karan Chatrath - 2 weeks, 6 days ago

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@Steven Chase @Karan Chatrath Please sir can you help me.. Thanks in advance.

Talulah Riley - 3 weeks, 1 day ago

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I'm getting approximately 0.35 0.35 for Problem 23. Is that the correct answer?

Steven Chase - 3 weeks ago

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@Steven Chase YES\Huge YES

Talulah Riley - 3 weeks ago

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My answer for problem 13 is:

Fmin=mg49μ2+1F_{min} = \frac{mg}{\sqrt{49 \mu^2 + 1}}

Is it correct?

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath No. In the answer the coefficient of μ2\mu^{2} is 11 instead of 4949.

Talulah Riley - 2 weeks, 6 days ago

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I am managing to get that answer but somehow that result is not making sense to me.

Karan Chatrath - 2 weeks, 6 days ago

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Okay, I have now understood the problem. What I did is also not wrong. I analysed a slightly different case.

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath @Karan Chatrath Yeah Nice. Can you show the solution?
OR would you like to see my attempt?

Talulah Riley - 2 weeks, 6 days ago

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Problem 13 solution:

Consider the force FF to have a horizontal and vertical component FxF_x and FyF_y respectively. Drawing a free body diagram will give you the following equations:

N=FxN = F_x Fymgf=maF_y - mg - f = ma fR=2mR25αfR = \frac{2mR^2}{5} \alpha

Now for the sphere to just start rolling upwards, it must just start linearly accelerating. In the limiting case, this happens when a=0a=0 and f=μNf = \mu N. Note that aa and α\alpha are unrelated. I was initially treating this as a pure rolling problem, but the problem does not specify that. This is a case of 'not pure rolling'. So plugging this understanding into the second equation above gives:

Fy=μFx+mgF_y = \mu F_x + mg

Now, the resultant force is:

F=Fx2+Fy2F = \sqrt{F_x^2+F_y^2}

The force FF is minimum when the expression in the square root is minimum. Therefore, we have a constrained minimisation problem. We are required to minimise the following function GG:

G(Fx,Fy)=Fx2+Fy2G(F_x,F_y) = F_x^2+F_y^2

Such that:

Fy=μFx+mgF_y = \mu F_x + mg

Solving this optimisation problem is something you should try yourself. The answer comes out to be:

Fmin=mgμ2+1F_{min} = \frac{mg}{\sqrt{\mu^2 + 1}}

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath Thanks for the solution.
I am trying to understand it. By the way ,the last 3rd step of your solution seems me incorrect, according to dimension. Isn't it.?

Talulah Riley - 2 weeks, 6 days ago

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I have used α\alpha for angular acceleration and aa for linear acceleration. Both symbols look similar.

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath @Karan Chatrath LAST 3rd step sir?

Talulah Riley - 2 weeks, 6 days ago

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@Karan Chatrath @Karan Chatrath g is a gravitational acceleration and F is a force.?

Talulah Riley - 2 weeks, 6 days ago

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@Talulah Riley Ah, good catch! I meant to define a function GG of FxF_x and FyF_y. I have modified the solution.

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath i didn't understand how did you optimised that at the last. Please elaborate the solution. Thanks in advance.

Talulah Riley - 2 weeks, 6 days ago

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Maxima and minima problem. Find the minimum value of the function:

G=Fx2+Fy2G = F_x^2 + F_y^2

Such that: Fy=μFx+mgF_y = \mu F_x + mg

Write FyF_y in terms of FxF_x and you have a function of FxF_x only. Do you know the next step?

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath @Karan Chatrath Yes i got it. The next step is differentiation and then just squaring.
But the most important thing is the moment of inertia is not used in this problem which seems me bit surprising.

Talulah Riley - 2 weeks, 6 days ago

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@Talulah Riley I used moment of inertia when I treated this problem as a case of pure rolling. That is why I got a factor of 49 multiplied with μ2\mu^2. But since it is not a pure rolling case, aRαa \ne R\alpha and therefore, the moment of inertia dependence is lost.

Karan Chatrath - 2 weeks, 6 days ago

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@Karan Chatrath @Karan Chatrath Exactly. It means friction is acting in the ball. Thanks By the way try the 14th question.

Talulah Riley - 2 weeks, 6 days ago

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@Karan Chatrath @Karan Chatrath i didn't understand the solution of this problem
Can you post a very explanatory solution.
After reading word rolling, my brain stops working . :)

Talulah Riley - 2 weeks, 6 days ago

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