What is the tension in the string at the top of the ball's path?

AI Thread Summary
The discussion focuses on calculating the tension in a string when a ball is at the top of its vertical circular path. The ball has a mass of 0.300 kg, a radius of 60.0 cm, and a speed of 4.00 m/s. Initially, the user incorrectly calculated the gravitational force (mg) as 5.88 N instead of the correct value of 2.94 N. After correcting this mistake, the tension in the string was recalculated to be 5.06 N. The importance of accurately calculating gravitational force in physics problems is emphasized.
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Homework Statement



Picture: http://img151.imageshack.us/img151/9393/physicszn8.jpg

A ball on the end of a string is cleverly revolved at a uniform rate in a vertical circle of radius 60.0 cm, as shown in Fig. 5-33. Its speed is 4.00 m/s and its mass is 0.300 kg.

(a) Calculate the tension in the string when the ball is at the top of its path.


Homework Equations



EF = ma


The Attempt at a Solution



EF = ma
Ft + Fg = ma
Ft = mac - Fg

Ft = ((mv^2)/r)) + mg
Ft = 8 - 5.88
Ft = 2.12

This answer is incorrect.
 
Last edited by a moderator:
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Recalculate mg.

;)
 
thejinx0r said:
Recalculate mg.

;)

Wow, I made a huge mistake... I had .6m * 9.8m/s^2 for mg... lol

New MG

0.300kg * 9.8m/sec^2 = 2.94.

8 - 2.94 = 5.06 N.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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