Angular Velocity and Tension Around a Fixed Cylinder

In summary, a particle of mass m is attached to a fixed vertical cylinder of radius a by a light string. The motion of the particle is in the horizontal plane and disregards gravity. At a distance b from the point of contact, the angular velocity of the string is w0. When the string turns through an additional angle theta, the angular velocity becomes w=w0/(1-(a/b)theta) and the tension in the string is T=m*b*w0*w. It is important to note that while angular momentum is not conserved due to a torque acting on the particle, the force on the particle from the string is always perpendicular to the velocity vector, allowing for a simple solution to be found. The assumption that the velocity
  • #1

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.
 
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  • #2
I think the velocity remains constant. Kinetic energy constant.
At least, that assumption leads to the given answer!
 
  • #3


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
 

1. What is angular velocity?

Angular velocity is a measure of the rate at which an object rotates around a fixed point. It is typically measured in radians per second or degrees per second.

2. How does angular velocity relate to rotational motion?

Angular velocity is a key component of rotational motion, as it describes how quickly an object is rotating around a fixed point. It is one of the factors that determines the amount of torque and centripetal force acting on an object.

3. What is tension in the context of a fixed cylinder?

In the context of a fixed cylinder, tension refers to the force that is exerted on the cylinder by an object or system that is attached to it. This force can be caused by the weight of the object, or by external forces acting on the object.

4. How does tension affect the angular velocity of a fixed cylinder?

Tension can affect the angular velocity of a fixed cylinder by providing a torque that causes the cylinder to rotate. The magnitude and direction of the tension force can impact the angular velocity, as well as the mass and distribution of mass of the object or system attached to the cylinder.

5. What factors can affect the tension and angular velocity of a fixed cylinder?

Several factors can affect the tension and angular velocity of a fixed cylinder, including the mass and distribution of mass of the object attached to the cylinder, the magnitude and direction of external forces acting on the object, and the length and stiffness of the connection between the object and the cylinder.

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