Identifying the forces acting on masses in uniform circular motion.

  • Thread starter deancodemo
  • Start date
  • #1
20
0

Homework Statement


Two particles of masses 2kg and 1kg are attached to a light string at distances 0.5m and 1m respectively from one fixed end. Given that the string rotates in a horizontal plane at 5 revolutions/second, find the tensions in both portions of the string.


Homework Equations





The Attempt at a Solution


First I convert the angular velocity from rev/sec to rad/sec.
[tex]\omega = 5[/tex] rev/sec
[tex]\omega = 10\pi[/tex] rad/sec

Now, let the 2kg mass be P, the 1kg mass be Q and the fixed point be O. I understand that the tension in PQ < tension in OP. Actually, the tension in OP should equal the centripetal force acting on P plus the tension in PQ. Right?
I get confused when identifying the forces acting on P. By using R = ma:
[tex]R = ma[/tex]
(Tension in OP) + (Tension in PQ) [tex]= m r \omega^2[/tex]
(Tension in OP) = [tex]m r \omega^2 - [/tex] (Tension in PQ)

This is incorrect!

The only solution would be:
[tex]R = ma[/tex]
(Tension in OP) - (Tension in PQ) [tex]= m r \omega^2[/tex]
(Tension in OP) = [tex]m r \omega^2 + [/tex] (Tension in PQ)

Which makes sense, but this means that the tension forces acting on P in the different sections of string act in opposite directions. Is that right?
 
Last edited:

Answers and Replies

  • #2
674
2
Draw a free body diagram of the forces acting on the mass at P. You will see both ropes on either side of the mass pulling in opposite directions. That is why your 2nd set of equations is correct.
 

Related Threads on Identifying the forces acting on masses in uniform circular motion.

  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
3K
Replies
5
Views
2K
Replies
5
Views
1K
Replies
1
Views
5K
  • Last Post
Replies
4
Views
6K
Replies
3
Views
1K
Replies
9
Views
1K
Replies
4
Views
1K
Top