# Homework Help: Find the initial acceleration of the block

1. Sep 17, 2015

### Vibhor

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

2. Relevant equations

3. The attempt at a solution

The question says block B has negligible mass ,which means particle will be effectively in free fall i.e acceleration would be 'g' . But then it is given a horizontal velocity i.e it has centripetal acceleration ac about the pulley v2/r .

The forces acting on the particle are Mg and tension T .

So, Mg -T = M(√(g2+ac2))

I am really clueless with this problem .

Please help me .

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2. Sep 17, 2015

### Orodruin

Staff Emeritus
If the mass of B is negligible, how is B going to affect the movement of A? (I.e., what is the tension in the string going to be?)

3. Sep 17, 2015

### Vibhor

B would not affect the movement of A . The tension would be zero ??

4. Sep 17, 2015

### Orodruin

Staff Emeritus
Yes, so how can you use this to find the acceleration of B?

5. Sep 17, 2015

### Vibhor

Acceleration of B = acceleration of the particle = √(g2+ac2) = 2√65 ??

Last edited: Sep 17, 2015
6. Sep 17, 2015

### Orodruin

Staff Emeritus
No, the acceleration of A is not the acceleration of B. Furthermore, the acceleration of A is just g in the down direction. It is also unclear why you would add the accelerations in quadrature.

I suggest that you think about how the motion of B relates to the motion of A. The expression v^2/r will appear, but not as you have stated.

7. Sep 17, 2015

### BvU

You have an idea about ac ? Does the particle follow a circular path (in view of what you said in post #3 ?)

8. Sep 17, 2015

### Staff: Mentor

After the particle has moved, if it has moved horizontally x and has moved down vertically y, what is its distance from the pulley in terms of H, x, and y?

Chet

9. Sep 17, 2015

### Vibhor

I agree,what I did doesn't make sense.But then if acceleration of A is just g , isn't A undergoing circular motion about the pulley ?? Why don't we account for v^2/r in the net acceleration of A ??

Sorry, I am bereft of any more ideas .

10. Sep 17, 2015

### Vibhor

$2H+\sqrt{x^2+y^2}$

11. Sep 17, 2015

### Staff: Mentor

Not exactly. Think Pythagorean theorem.

Chet

12. Sep 17, 2015

### Vibhor

2H is the initial distance of A from pulley . It moves an additional distance $\sqrt{x^2+y^2}$ . Isn't it now at a distance $2H+\sqrt{x^2+y^2}$ from the pulley ??

13. Sep 17, 2015

### Staff: Mentor

No. One leg of the big right triangle is (2H+y) and the other leg of the triangle is x. The distance from the pulley to the particle is the hypotenuse of the triangle. Try again.

Chet

14. Sep 17, 2015

### Vibhor

Should it be $\sqrt{x^2+(2H+y)^2}$ ??

15. Sep 17, 2015

### Staff: Mentor

Yes. Good.

Now how much has the distance increased relative to what it was at time zero? Given that the string is of constant total length, in terms of x, y, and H, how much has mass B moved to the right?

Chet

16. Sep 17, 2015

### Vibhor

$\sqrt{x^2+(2H+y)^2} - 2H$

17. Sep 17, 2015

### Staff: Mentor

Good. Now what is x as a function of t and y as a function of t? Substitute these expressions into the above equation for the distance d that the mass B has moved to the right.

Chet

18. Sep 17, 2015

### Vibhor

$x=vt$ and $y=\frac{1}{2}gt^2$

$d = \sqrt{(vt)^2+(2H+(\frac{1}{2}gt^2))^2} - 2H$

19. Sep 17, 2015

### Staff: Mentor

OK. Now next use the expression for v given in the problem statement and expand out the expression in parenthesis. Let's see what you get.

Chet

20. Sep 17, 2015

### Vibhor

$d = \sqrt{\frac{1}{4}g^2t^4+10Hgt^2+4H^2} - 2H$

21. Sep 17, 2015

### Staff: Mentor

Good. Now, if I factor out 4H2 from under the square root sign, I get:
$$d =2H \sqrt{1+\frac{5}{2}\frac{gt^2}{H}+\frac{1}{16}\left(\frac{gt^2}{H}\right)^2} - 2H$$
Are you familiar with how to approximate $\sqrt{1+x}$ for small values of x? Making this approximation would be very convenient (but not mandatory) at this point in our analysis, since we are interested in the behavior at very short times.

Chet

22. Sep 17, 2015

### Vibhor

$\sqrt{1+x} ≈ 1+\frac{1}{2}x$

$$d =2H\left(1 +\frac{1}{2} \left(\frac{5}{2}\frac{gt^2}{H}+\frac{1}{16}\left(\frac{gt^2}{H}\right)^2\right)\right)- 2H$$

$$d =H \left(\frac{5}{2}\frac{gt^2}{H}+\frac{1}{16}\left(\frac{gt^2}{H}\right)^2\right)$$

$$d =\frac{5}{2}(gt^2)+\frac{1}{16}\left(\frac{g^2t^4}{H}\right)$$

$$\ddot{d}(t=0) = 5g = 50ms^{-2}$$

Is this what I should get ??

Last edited: Sep 17, 2015
23. Sep 17, 2015

### Staff: Mentor

Sure. Very nice job.

Chet

24. Sep 17, 2015

### Vibhor

The answer given is 20ms-2 .

Last edited: Sep 17, 2015
25. Sep 17, 2015

### Staff: Mentor

Check over what we've done and see if we made a mistake in "arithmetic." Our methodology is definitely correct.

Chet

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