Linear Momentum and the center of mass

  • #1


On a horizontal air track, a glider of mass m carries a post shaped like an inverted "L". The post supports a small dense sphere, also of mass m, hanging just above the top of the glider on a cord of length L. The glider and sphere are initially at rest with the cord vertical. A constant horizontal force of magnitude F is applied to the glider, moving it through displacement x1; then the force is removed. During the time interval when the force is applied, the sphere moves through a displacement with a horizontal component of x2.
(a) Find the horizontal component of the velocity of the center of mass of the glider-sphere system when the force is removed.
(b) After the force is removed, the glider continues to move on the track and the sphere swings back and forth, both without friction. Find an expression for the largest angle the cord makes with the vertical.

At first, I naively took the total energy of the system to be

E = F*x1 = 0.5(2m)v^2 ,

Then it occurred to me that the displacement x2 of the sphere does some work as well, due to the presence of the fictitious force (measured in the glider's frame of reference). This also means that it rises to a height above it's original position, and work done against gravity becomes mgh. What really killed me was when I realized that the shift in the position has to affect the system's motion in some way, since the center of mass remains constant.

Would really appreciate it if someone enlightens me, for I now have no idea where to start.
Thanks in advance. :)

Ans: (a) V_CM = √(F(x1 + x2)/(2m));
(b) θ = arcos(1 - F(x1 + x2)/(2mgL))

Source: "Physics for Scientists and Engineers with Modern Physics" by Jewett and Serway, 8th edition, page 276
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  • #2
Start with the work-energy theorem, ##\Delta K=W_{net}.## Assume that, just before the force is removed, the cord makes angle ##\alpha## relative to the vertical and the hanging mass is at rest relative to the cart. Then, the horizontal component of the velocity of the CM is the common final speed ##v_f## at the moment the force is removed. That component is conserved because there are no external horizontal forces acting on the system when ##F## is removed.
1. The work done by gravity on the ball is ##W_g=-mgL(1-\cos\alpha)##.
2. The work done by the external force on the CM is ##W_F=F~x_1##.
3. The change in kinetic energy of the CM is ##\Delta K=\frac{1}{2}(2m)v_f^2=mv_f^2.##
Now from a free body diagram of the hanging mass, the vertical component of the tension must match the weight and the horizontal component must provide its acceleration, ##T \cos\alpha=mg## and ##T \sin\alpha = ma.##
Dividing the second equation by the first gives ##\tan\alpha=\dfrac{a}{g}.##
Now the acceleration of the system is ##a=\dfrac{F}{2m} ~\rightarrow~\tan\alpha=\dfrac{F}{2mg}##.
Note that $$\cos\alpha=\frac{1}{\sqrt{1+\tan^2\alpha}}=\frac{2mg}{\sqrt{(2mg)^2+F^2}}$$We now put all this in the equation for the work-energy theorem to get$$mv_f^2=F~x_1-mgL\left(1-\frac{2mg}{\sqrt{(2mg)^2+F^2}}\right)$$
Answer to part (a)
$$v_f=\left[ \frac{F~x_1}{m}-gL\left(1-\frac{2mg}{\sqrt{(2mg)^2+F^2}}\right)\right]^{1/2.}$$
The answer to part (b) has already been found
$$\alpha=\arctan \left(\frac{F}{2m}\right).$$
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