Circular motion (2 particles on a string)

[Geometor]

Homework Statement

A light inextensible string of length 0.6m has one end fixed to a point A on a smooth horizontal plane. The other end of the string is attactched to a particle B, of mass 0.4kg. A particle P of mass 0.1kg is attatched to the mid-point of the string. The line APB is straight and rotation continues at 2 rad s^-1 on the surface of the plane.

1) Calculate the tension in the section of the string AP.
2) Calculate the tension in the section of the string BP.

Homework Equations

Centripetal force = mass x radius x (angular velocity)^2

The Attempt at a Solution

I imagined that the tension in AP would simply be:
0.1 x 0.3 x 2^2 = 0.12N

however the answer is given as 1.08N. So my question really is how does two particles on the same string affect the tension?

I placed in q.2 for curiosity.

Thanks.

 

[PhanthomJay]

You must draw free body diagrams . You should first start with question 2 to determine the string tension BP by looking at particle B alone. Then look at particle P alone. There are 2 tension forces acting on particle P.

 

[Geometor]

Excuse the poor diagram but:

A--->---<---P--->---<---B

where > and < indicates direction of the force
from left to right forces are: T1, T1, T2, T2

Considering B alone, resolving horizontally:
T(at B) = 0.4 x 0.6 x 2^2 = 0.96N

Considering P alone, resolving horizontally:
T(at P) = 0.1 x 0.3 x 2^2 = 0.12N

So I suppose this says that the resultant force on P is 0.12N.
So if we take left as positive at P,

T1 - T2 = 0.12N
T1 = 0.12 + 0.96
= 1.08N

I believe this is the correct line of thinking?

Also, if 2 particles hang on a string (light, inextensible) vertically, one above another. Will the tension in the top string be the sum of the weight of the two particles?

Thank you.

 

[PhanthomJay]

Excuse the poor diagram but:

A--->---<---P--->---<---B

where > and < indicates direction of the force
from left to right forces are: T1, T1, T2, T2

Considering B alone, resolving horizontally:
T(at B) = 0.4 x 0.6 x 2^2 = 0.96N

The net (resultant) horizontal force at B is 0.96N; since the only horizontal force acting at B is T1, then T1 = 0.96N

Considering P alone, resolving horizontally:
T(at P) = 0.1 x 0.3 x 2^2 = 0.12N

again, this is the resultant horizonatal force at P

So I suppose this says that the resultant force on P is 0.12N.

yes

So if we take left as positive at P,

T1 - T2 = 0.12N
T1 = 0.12 + 0.96
= 1.08N

I believe this is the correct line of thinking?

yes

Also, if 2 particles hang on a string (light, inextensible) vertically, one above another. Will the tension in the top string be the sum of the weight of the two particles?

yes

 

[rwisz]

I would like to clarify that the tension in a string is a result of forces acting on it. In this case, we have two particles attached to the string, which means that the string is being pulled in two directions. Therefore, the tension in the string will be affected by both particles.

To calculate the tension in the section of the string AP, we need to consider the forces acting on particle P. We know that the particle is moving in a circular motion, which means there must be a centripetal force acting on it. This force is provided by the tension in the string. Using the equation for centripetal force, we can calculate the tension in the string AP as:

Tension in AP = (mass of particle P) x (radius of circular motion) x (angular velocity)^2

= (0.1 kg) x (0.3 m) x (2 rad/s)^2

= 0.12 N

Therefore, your calculation for the tension in AP is correct.

Moving on to the tension in the section of the string BP, we need to consider the forces acting on particle B. Since particle B is attached to the string, it will also experience a tension force. However, in this case, the string is not only providing a centripetal force but also supporting the weight of particle B. This means that the tension in the string BP will be equal to the weight of particle B plus the centripetal force. We can calculate this as:

Tension in BP = (mass of particle B) x (gravitational acceleration) + (mass of particle B) x (radius of circular motion) x (angular velocity)^2

= (0.4 kg) x (9.8 m/s^2) + (0.4 kg) x (0.3 m) x (2 rad/s)^2

= 0.4 N + 0.12 N = 0.52 N

Therefore, the answer given for the tension in BP (1.08 N) seems to be incorrect. Using the given values, the correct value for the tension in BP should be 0.52 N.

In conclusion, when dealing with circular motion involving multiple particles on a string, we need to consider the forces acting on each particle separately to accurately calculate the tension in the string. I hope this helps clarify any confusion.