Two balls are connected to a string and execute circular motion in a horizontal plane as shown below. The outer ball has mass M1 and the inner ball has a mass M2. Equal lengths of string (L) are attached to M2.

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/hwb/07/01/two_balls_horiz.gif

The tension in the string attached to both balls is less than the tension in the string attached only to M2.

TRUE
FALSE

What is T1, the magnitude of the tension in the string attached to both balls? Assume M1 = 0.3kg, M2 = 0.6kg, L = 0.4m, and v2 = speed of M2 = 0.8m/s.

T1 = 0.24N
T1 = 0.48N
T1 = 0.96N
T1 = 2.94N
T1 = 3.90N

What are the forces implied in each string?

The centripetal force can be written as:

F_c = \frac{mv^2}{r} = mr \omega^2 = mv\omega

Since omega is constant while v is dependent on M1 and M2, you take the second form.

F_c = mr\omega^2

What can you say about the force in the strings now?

What I think is that for the first 1, because there are more mass, so there would be more tension than the string that has only one mass, so I think the first statement is "FALSE"

For the second question, I don't know where to start

Yes, that's correct :)

2. As I told you, omega is constant for both balls. So, find omega first;

v = r\omega

r is the length L, you know v.

Then, find the centripetal force that M1 needs if it was alone using

F_1 = mr\omega^2 = M_1(2L)\omega^2

In fact, in the string connecting the centre provides the centripetal force of ball 1 and ball 2. The string connection both balls provides the centripetal force of ball 1 only.

Since you know the centripetal force holding ball 1, you know the tension in the string connecting both balls.

Post what you get! :)

Is mass one and mass two moving at the same velocity, why's that?

So the magnitude should be 0.96N? And I don't quite get what you mean by this statement

"In fact, in the string connecting the centre provides the centripetal force of ball 1 and ball 2. The string connection both balls provides the centripetal force of ball 1 only.

Since you know the centripetal force holding ball 1, you know the tension in the string connecting both balls."

Both balls go at the same angular velocity (\omega). They do a complete turn in the same time. However, they don't have the same linear speed (v). That's why I use the second formula instead of the other two.

That said, let's work out the angular velocity of the balls.

v = r\omega

\omega = \frac{0.8}{0.4} = 2 rad/s

Let's call the tension in the string above ball 2, S2 and that in the string connecting both balls S1.

As S2 holds both balls, we get the tension in S2 as:

F_{S2} = M_2r\omega^2 + M_12r\omega^2 = (2M_1 + M_2)r\omega^2 = (2(0.3) + 0.6)(0.4)(2^2) = 1.92 N

If S2 held ball 1 only, it would have a tension of:

F = mr\omega^2 = (0.6)(0.4)(2^2) = 0.96 N

So, Tension in S2 is the difference, or (1.92 - 0.96) = 0.96 N

Yes, it's good :)