Angular Velocity and Tension Around a Fixed Cylinder

[s_stylie0728]

Homework Statement

A particle of mass m at the end of a light string wraps itself about a fixed vertical cylinder of radius a. All the motion is in the horizontal plane (disregard gravity). The angular velocity of the cord is w0 when the distance from the particle to the point of contact of the string and cylinder is b. Find the angular velocity and tension in the string after the cord has turned through an additional angle theta.

Hints from my professor:
-There is a torque acting on m so angular momentum is not conserved. But something even more simple is.
-The force on m due to the string is always perpendicular to the velocity vector.

Homework Equations

T=mv^2/r
w= d(theta)/dt
w=vsin(theta)/r
?

The answer given is:
w=w0/(1-(a/b)theta), T=m*b*w0*w

The Attempt at a Solution

I saw that v*sin(theta) = w0 so I got as far as w=w0/r but that's it. And I'm confused in the tension equation as to how one would obtain a as b*w0*w.

 

[Delphi51]

I think the velocity remains constant. Kinetic energy constant.
At least, that assumption leads to the given answer!

 

[nlightner]

I would first clarify the problem statement to make sure I understand the setup correctly. It seems that the particle is attached to a string, which is wrapped around a fixed cylinder. The particle is moving in a horizontal plane and the only force acting on it is the tension in the string. The problem is asking for the angular velocity and tension in the string after the particle has turned through an additional angle theta.

To solve this problem, we need to apply the principles of rotational dynamics. The first hint from the professor suggests that angular momentum is not conserved, but something simpler is. This "something simpler" is the conservation of energy. Since there is no external torque acting on the system, the total energy remains constant.

We can use this principle to find the angular velocity after the particle has turned through an additional angle theta. At the initial position, the particle has a kinetic energy of 1/2 mv^2 and a potential energy of mgb, where g is the acceleration due to gravity. At the final position, the particle has a kinetic energy of 1/2 m(v+w)^2 and a potential energy of mg(b-a). Since the total energy is conserved, we can equate these two expressions and solve for w:

1/2 mv^2 + mgb = 1/2 m(v+w)^2 + mg(b-a)
1/2 mv^2 + mgb = 1/2 mv^2 + mvw + 1/2 mw^2 + mgb - mga
1/2 mw^2 = m(vw - ga)
w = (2vw - 2ga)/m

Substituting in the expression for v*sin(theta) = w0, we get:

w = (2w0*sin(theta) - 2ga)/m

This is the angular velocity of the particle after it has turned through an additional angle theta. To find the tension in the string, we can use the second hint from the professor, which states that the force on the particle due to the string is always perpendicular to the velocity vector. This means that the work done by the tension is zero and therefore the tension does not contribute to the change in kinetic energy. Hence, the tension remains constant and is equal to the centripetal force required to keep the particle moving in a circle. Using the equation for centripetal force, we get:

T = mv^2/r = m(w