Why Does Cutting the String Change the Tension Ratio in a Pendulum?

AI Thread Summary
Cutting the string attached to the wall changes the forces acting on the ball, leading to different tension ratios in the supporting string. Initially, the ball is in equilibrium, and the tension is calculated as T1 = mg/cos(theta). Once the wall string is cut, the ball swings as a pendulum, and the tension during this motion is T2 = mg cos(theta). The ratio of the tensions before and after cutting the string is T1/T2 = 1/cos^2(theta), indicating that the tension increases significantly as the pendulum swings. This change is due to the different forces acting on the ball in equilibrium versus during its motion.
daveed
Messages
138
Reaction score
0
SO.

a ball is suspended by a string from the ceiling, and held at angle theta to the vertical by another string that is attached to the wall nearby... the string attched to the wall is cut and the ball swings like a pendulum, at the farthest it is theta from the vertical on both sides.

so. the book asks what is the ratio of the tension in the rope holding the rope to the ceiling before the other rope(attached to the wall) was cut, to when the ball is swinging and is at its farthest from the vertical(forming angle theta to it)

i thought it should be one to one, because the tension*sin(theta) would be supporting the ball in both cases, but it turns out the answer is cos^2(theta)

any ideas why?
 
Physics news on Phys.org
Check your vector diagram.
 
\begin{align*}\\<br /> Case\ 1 :\\<br /> Particle\ is\ in\ equilibrium. \\<br /> \sum \vec{F}=0\\<br /> T_1cos\theta + (-mg) = 0\\<br /> T_1=\frac{mg}{cos\theta}\\<br /> Case\ 2:\\<br /> Centripedal\ acceleration :\\<br /> T_2+(-mgcos\theta)=0\ since\ v=0\\<br /> T_2=mgcos\theta\\<br /> \frac{T_1}{T_2}=\frac{1}{cos^2\theta}\\<br /> Why\ the\ reciprocal\ ?\\<br /> \end{align}
 

Attachments

  • Picture.GIF
    Picture.GIF
    2.9 KB · Views: 547
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top