Find the initial acceleration of the block

[Vibhor]

Homework Statement

Homework Equations

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 a_c about the pulley v²/r .

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

So, Mg -T = M(√(g²+a_c²))

I am really clueless with this problem .

Please help me .

 

[Orodruin]

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?)

 

[Vibhor]

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?)

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

 

[Orodruin]

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

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

 

[Vibhor]

Acceleration of B = acceleration of the particle = √(g²+a_c²) = 2√65 ??

 

[Orodruin]

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.

 

[BvU]

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

 

[Chestermiller]

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

 

[Vibhor]

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 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 ??

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.

Sorry, I am bereft of any more ideas .

 

[Vibhor]

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

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

 

[Chestermiller]

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

Not exactly. Think Pythagorean theorem.

Chet

 

[Vibhor]

Not exactly. Think Pythagorean theorem.

Chet

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 ??

 

[Chestermiller]

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 ??

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

 

[Vibhor]

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

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

 

[Chestermiller]

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

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

 

[Vibhor]

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

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

 

[Chestermiller]

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

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

 

[Vibhor]

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

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

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

 

[Chestermiller]

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

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

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

 

[Vibhor]

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

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

 

[Chestermiller]

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

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

 

[Vibhor]

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

$\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 ??

 

[Chestermiller]

Sure. Very nice job.

Chet

 

[Vibhor]

The answer given is 20ms^-2 .

 

[Chestermiller]

The answer given at the back is 20ms^-2 .

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

Chet

 

[TSny]

I also get an answer of 5g using polar coordinates.

 

[Chestermiller]

I also get an answer of 5g using polar coordinates.

That clinches it.

 

[Vibhor]

I also get an answer of 5g using polar coordinates.

Thanks for confirming the result , Sir . Could you please show me how you worked this problem using polar coordinates .

 

[TSny]

Could you please show me how you worked this problem using polar coordinates .

Let the origin be at the pulley and use polar coordinates $r$ and $\theta$ for the position of the ball. In polar coordinates the radial component of acceleration is $a_r = \ddot{r} – r\dot{\theta}^2$ (See for example https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=dy&chap_sec=01.6&page=theory )

Which of the two terms on the right represents the acceleration of the block?

Just after the ball is given its initial velocity $v_0$, what is the value of $\dot{\theta}$?

Just after the ball is given its initial velocity, the radial direction is in the vertical direction (downward). Since the tension is zero, you know the value of the acceleration in the downward direction. Thus, you know the value of $a_r$ at the initial time.

So, you have everything to get the acceleration of the block

 

[Vibhor]

Let the origin be at the pulley and use polar coordinates $r$ and $\theta$ for the position of the ball. In polar coordinates the radial component of acceleration is $a_r = \ddot{r} – r\dot{\theta}^2$ (See for example https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=dy&chap_sec=01.6&page=theory )

Which of the two terms on the right represents the acceleration of the block?

$\ddot{r}$

Just after the ball is given its initial velocity $v_0$, what is the value of $\dot{\theta}$?

$\dot{\theta}=\frac{v_0}{{2H}}$

Just after the ball is given its initial velocity, the radial direction is in the vertical direction (downward). Since the tension is zero, you know the value of the acceleration in the downward direction. Thus, you know the value of $a_r$ at the initial time.

So, you have everything to get the acceleration of the block

acceleration of the block = $\ddot{r} = a_r + r\dot{\theta}^2 = g + \frac{{v_0}^2}{2H} = g+4g = 5g$

Does that look alright ??