Calculating Tension in a Rope: A 2kg Stone and 1m Radius

AI Thread Summary
The discussion focuses on calculating the tension in a rope attached to a 2kg stone moving in a circular path with a radius of 1 meter and a velocity of 4 m/s. For the top point of the circle, the tension is calculated as 12.4 N, while at the bottom point, the tension increases to 51.6 N. The calculations involve using the formulas for centripetal force and gravitational force. Participants confirm that the solutions provided are accurate. The discussion effectively clarifies the tension dynamics in circular motion.
Atilla1982
Messages
18
Reaction score
0
A stone with mass 2,0kg is tied to a rope with radius 1 meter to the circle center, and has the velocity 4 m/s.

a) What's the tension on the rope on the top point of the circle?

b) What's the tension on the rope on the bottom point of the circle?


Here's my attempt to solve these.

a) Fc = mv^2/r Fg= mg Ft= the force applied on the rope

Fc=Ft + Fg
Ft=Fc - Fg
Ft=mv^2/r - mg
Ft=2*4^2/1 - 2*9,8
Ft=12,4 N

b)
Fc=Ft - Fg
Ft=Fc + Fg
Ft=mv^2/r + mg
Ft=2*4^2/1 + 2*9,8
Ft=51,6 N

Can anyone please verify that this is correct?
 
Physics news on Phys.org
Both of your solutions are correct.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top