# Mechanics Doubt HELP!

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

Note by Talulah Riley
5 months ago

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Is the answer to problem 27 $1.5 mg$?

- 5 months ago

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

- 5 months ago

I will post a note later.

- 5 months ago

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

- 5 months ago

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

- 5 months ago

@Steven Chase $\Huge YES$

- 5 months ago

My answer for problem 13 is:

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

Is it correct?

- 5 months ago

@Karan Chatrath No. In the answer the coefficient of $\mu^{2}$ is $1$ instead of $49$.

- 5 months ago

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

- 5 months ago

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

- 5 months ago

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

- 5 months ago

Problem 13 solution:

Consider the force $F$ to have a horizontal and vertical component $F_x$ and $F_y$ respectively. Drawing a free body diagram will give you the following equations:

$N = F_x$ $F_y - mg - f = ma$ $fR = \frac{2mR^2}{5} \alpha$

Now for the sphere to just start rolling upwards, it must just start linearly accelerating. In the limiting case, this happens when $a=0$ and $f = \mu N$. Note that $a$ and $\alpha$ 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:

$F_y = \mu F_x + mg$

Now, the resultant force is:

$F = \sqrt{F_x^2+F_y^2}$

The force $F$ 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 $G$:

$G(F_x,F_y) = F_x^2+F_y^2$

Such that:

$F_y = \mu F_x + mg$

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

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

- 5 months ago

@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.?

- 5 months ago

I have used $\alpha$ for angular acceleration and $a$ for linear acceleration. Both symbols look similar.

- 5 months ago

@Karan Chatrath LAST 3rd step sir?

- 5 months ago

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

- 5 months ago

Ah, good catch! I meant to define a function $G$ of $F_x$ and $F_y$. I have modified the solution.

- 5 months ago

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

- 5 months ago

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

$G = F_x^2 + F_y^2$

Such that: $F_y = \mu F_x + mg$

Write $F_y$ in terms of $F_x$ and you have a function of $F_x$ only. Do you know the next step?

- 5 months ago

@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.

- 5 months ago

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 $\mu^2$. But since it is not a pure rolling case, $a \ne R\alpha$ and therefore, the moment of inertia dependence is lost.

- 5 months ago

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

- 5 months ago

@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 . :)

- 5 months ago