How can the motion of point A be determined in this system?

[Vibhor]

By the way, this is likely to get a little tricky. I don't think the person who conceived of this problem realized how complicated it is.

Chet

I think , I should leave this problem as it has already taken up so much of my time . I had been thinking about this problem for past few days .After some real hard thinking I decided to post it here . But after so many posts , there hasn't been any progress .

Thank you Sir for your valuable time . I really appreciate your help .

 

[Chestermiller]

I think , I should leave this problem as it has already taken up so much of my time . I had been thinking about this problem for past few days .After some real hard thinking I decided to post it here . But after so many posts , there hasn't been any progress .

Thank you Sir for your valuable time . I really appreciate your help .

I have one more thing to say before stopping. Since the horizontal velocity of A is zero at time zero, then, as reckoned along a line from the connection point at the ceiling to point A, the centripetal term in the expression for radial acceleration of A toward the connection point at the ceiling is zero. And since the wire is presumed inextensible, the second derivative of the distance between the connection point at the ceiling and point A is also zero. The sum of these two terms is equal to the vertical component of the acceleration of point A at time zero. So, at time zero, the vertical component of acceleration of point A is equal to zero.

Chet

 

[haruspex]

@Vibhor, @Chestermiller: you seem to have been at pains to show that there is no vertical acceleration at t=0. The question does not offer nonzero vertical acceleration as an option anyway, so I would think the question setter accepts it is obvious there is none. There cannot be a vertically downward acceleration, and the tension in the string will only be sufficient to prevent that, and there is no other force to create an upward acceleration.

Thus, I read the question as purely a matter of deciding whether there is a nonzero horizontal acceleration at t=0. This is not as completely trivial as two other posters on the thread thought, but it can be settled with a fairly straightforward argument, making the question and solution satisfying.

However, this reading is belied by the wording of the question. It asks about a short time after the B thread combusts, not t=0. Not only does that make it far too obvious that there will be a horizontal acceleration, there will also be a vertical acceleration, making the offered answers both incorrect. Therefore I believe the wording does not describe the intended problem. It should be asking about t=0. But that's just an opinion.

 

[Chestermiller]

@Vibhor, @Chestermiller: you seem to have been at pains to show that there is no vertical acceleration at t=0. The question does not offer nonzero vertical acceleration as an option anyway, so I would think the question setter accepts it is obvious there is none.

The fourth option in the problem statement says that "The acceleration of point A is horizontal." To me this means that you need to show that the horizontal acceleration is non-zero, and the vertical acceleration is zero.

There cannot be a vertically downward acceleration, and the tension in the string will only be sufficient to prevent that, and there is no other force to create an upward acceleration.

In post #32, I believe I presented a very convincing proof that the vertical acceleration of A is zero immediately after the cord is cut.

Thus, I read the question as purely a matter of deciding whether there is a nonzero horizontal acceleration at t=0. This is not as completely trivial as two other posters on the thread thought, but it can be settled with a fairly straightforward argument, making the question and solution satisfying.

Vibhor and I analyzed the problem mathematically, and let the math do the work for us to show that the horizontal acceleration is nonzero.

However, this reading is belied by the wording of the question. It asks about a short time after the B thread combusts, not t=0. Not only does that make it far too obvious that there will be a horizontal acceleration, there will also be a vertical acceleration, making the offered answers both incorrect. Therefore I believe the wording does not describe the intended problem. It should be asking about t=0. But that's just an opinion.

It's pretty clear to me (at least in my personal judgement) that they meant for you to assume that at time t = 0⁺, all the velocities are zero.

 

[haruspex]

The fourth option in the problem statement says that "The acceleration of point A is horizontal." To me this means that you need to show that the horizontal acceleration is non-zero, and the vertical acceleration is zero.

In post #32, I believe I presented a very convincing proof that the vertical acceleration of A is zero immediately after the cord is cut.

Vibhor and I analyzed the problem mathematically, and let the math do the work for us to show that the horizontal acceleration is nonzero.

It's pretty clear to me (at least in my personal judgement) that they meant for you to assume that at time t = 0⁺, all the velocities are zero.

Ok, thanks - I must have not followed your exchanges closely enough.

 

[ehild]

The answers become clear when we write the equations for the motion of the CM and point A in general. In this case, the string encloses the angle φ with the vertical while the beam encloses angle θ. Let be the length of the beam 2L, and r the length of the string.
x_A=rsin(φ), y_A=rcos(φ)
$\ddot x_A=−rsin(φ){\dot φ}^2+rcos(φ)\ddot φ$, $\ddot y_A=−rcos(φ){\dot φ}^2−rsin(φ)\ddot φ$
At t=0+, φ=0, $\dot φ=0$, so $\ddot x_A=r\ddot φ$, $\ddot y_A=0$, point A accelerates in horizontal direction.

$X_{CM}=L\sin(θ)+rsin(φ)$, $Y_{CM}=L\cos(θ)+r\cos(φ)$
$\ddot X_{CM}=-L\sin(θ){\dot θ}^2+L\cos(θ)\ddotθ-r\sin(φ){\dot φ}^2+r\cos(φ)\ddotφ$
$\ddot Y_{CM}=-L\cos(θ){\dot θ}^2-L\sin(θ)\ddot θ-r\cos(φ){\dot φ}^2-r\sin(φ)\ddotφ$
At t=0, $\ddot X_{CM}=L\cos(θ)\ddot θ+r\ddot φ$, $\ddot Y_{CM}=-L\cos(θ)\ddot θ$

For the translation of the CM:
$m\ddot X_{CM}=-T\sin (φ)$
$m\ddot Y_{CM}=-T\cos (φ)+mg$
The torque equation with respect to the CM:
$TLsin(θ-φ)=-\frac{1}{3}mL^2 \ddot θ$.
At t=0:
$\ddot X_{CM}=0$, $\ddot Y_{CM}=-T+mg$
$TLsin(θ)=-\frac{1}{3}mL^2 \ddot θ$.

T can be eliminated, $\ddot θ$and $\ddot φ$ determined for t=0.