Question

A rigid massless rod is rotated about one end in a horizontal circle. There is a load of mass m attached to the center of the rod, and one of mass M attached to the outer end of the rod. The inner section of the rod sustains a tension that is three times as great as the tension that the outer section sustains. The mass ration of m/M is?

Answer is 1

WHY?

Answer 1

Let me denote m as m1 and M as m2.

The basic formula that we know in circular motion is:

F=mv^2/r,

where F is the centripetal force, m the mass, v is the linear velocity, and r is the radius.

We also know that v=wr

where v is the linear speed, w (supposed to be omega) is the angular velocity, and r is the radius.

Substitute the value of v in our first formula:

F=m(wr)^2/r

F =mw^2r Formula 1)

In this problem, we also know that w which is the angular velocity is the same for m1 as for m2 because they are located along the same line which is rotating about one end of the rod.

We are also given the value of the tension in the inner section as 3x that of the outer section. Let's draw a simple free body diagram:

T1=3F2<----0-->T2=F2<---0

The first 0 is m1 and the second 0 is m2.

The diagram might get distorted when I send it, but let's just cross our fingers it comes out clear enough.

The net force acting on m1 is (3F2-T2). But T2=F2.

Therefore, net force on m1 =3F2-F2=2F2. Let's call

this net force as F1. Thus F1=2F2. Now we also know

that

F1=m1w^2r

based on formula 1) above. Now substitute the value of F1=2F2, and we get:

2F2=m1w^2r Equation 1)

And based on the same formula F=mw^2r:

F2=m2w^2(2r) Equation 2)

Remember that for m2 the radius is twice that of m1, which is a given condition in this problem.

Now divide equation 2 by equation 1:

F2=m2w^2(2r)

------------------

2F2=m1w^2r

F2,w^2, and r will cancel out leaving:

1/2=2m2/m1

Divide both sides by 2:

1/4=m2/m1

m2/m1=1/4

=> m1/m2 = 4

or,

m/M = 4

So the answer is 4 and not 11.