# A 4.12-kg object is attached to a vertical rod by ...

A 4.12-kg object is attached to a vertical rod by two strings, as in the figure below. The object rotates in a horizontal circle at constant speed 6.95 m/s . sorry figure wouldn't show up but it is set up such that the ball swings around the middle of the rod, the strings are 1.84 meters long each with one end attached to the ball and the other end to the ends of the rod. the rod is 3.2 meters long (a) Find the tension in the upper string. Take the free fall acceleration to be 9.8 Round your answer to three significant figures. N (b) Find the tension in the lower string. Round your answer to three significant figures. N

1. isthis how your image looks like? if so, the Tension for the firstone will be equal to centripetal force, which is equalto
The Tension of the lower string will simply be

OH, SORRY. THE BOTTOM MASS IS m2!!!
2. A slight twist to that in the textbook. The same methodis used by both situations:
Step 1: notice the shapes involved in the problem:
"I" above is meant to be theta. To determin r cut theresulting triangle in half and use trig to find r and theta
? =60.4
Step 2: Look at the Free Body diagram
W=mg
? Fy=ma -There is no accelleration in they-direction
i.)
? Fx=ma -There is centripital accelleration in thex-direction
ii.)
Now we have two equations for two unknowns.
From i.):
Put this equality into ii.)

Now just solve for Tu

Tu= 221.6 + 23.22=244.8N
TL is found this way aswell or one could just plug thisvalue into equations i, or ii. to get TL.
Thus:
from i.)
TL=198.4N