Identifying the forces acting on masses in uniform circular motion.

[deancodemo]

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.
$$\omega = 5$$ rev/sec
$$\omega = 10\pi$$ 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:
$$R = ma$$
(Tension in OP) + (Tension in PQ) $$= m r \omega^2$$
(Tension in OP) = $$m r \omega^2 -$$ (Tension in PQ)

This is incorrect!

The only solution would be:
$$R = ma$$
(Tension in OP) - (Tension in PQ) $$= m r \omega^2$$
(Tension in OP) = $$m r \omega^2 +$$ (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?

 

[nickjer]

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.

 

[nlightner]

I would like to clarify that in uniform circular motion, the net force acting on an object is always directed towards the center of the circle. In this case, the center of the circle is the fixed point O. Therefore, the only force acting on P is the tension in the string, which is directed towards O. This tension force is necessary to keep P moving in a circular path.

In this problem, the tension in the string at point P (closer to O) will be greater than the tension at point Q (further from O). This is because the centripetal force required to keep P in circular motion is greater due to its larger distance from the center of the circle.

So, the forces acting on P in uniform circular motion are the tension force towards O and the centripetal force towards the center of the circle. The tension force in the string can be broken down into two components: the tension in OP and the tension in PQ. The tension in OP is responsible for providing the necessary centripetal force, while the tension in PQ is responsible for balancing out the weight of P and keeping it in equilibrium.

Therefore, in this problem, the tension in OP can be calculated using the equation T = m r \omega^2, where m is the mass of P, r is the distance from P to O, and \omega is the angular velocity in rad/sec. The tension in PQ can be found by subtracting the tension in OP from the total tension in the string.

In summary, the forces acting on masses in uniform circular motion are the tension force towards the center of the circle and the centripetal force towards the center of the circle. These forces are necessary to maintain the circular motion of the objects.