What is the thread's length pulled and the variation factor of velocity?

• Icaro Amorim
In summary, the problem involves a ball being held by a thread and moving in a circular motion around a vertical axis. The thread is pulled up, changing the angle and velocity of the ball. The goal is to find the length of the thread pulled and the variation factor of velocity. Equations and attempts at solving the problem are given, including the suggestion to use conservation of angular momentum to solve for the unknowns.
Icaro Amorim

Homework Statement

Hello, guys. I'm not a native english speaker, so don't mind if I wrote something incorrectly (if you find any error or difficulty in what I wrote just inform me, please).

A little ball is hold by a thread of negligible mass which moves round a vertical axis with constant velocity. It keeps a distance of 0,5m from the axis when the anglo θ is equal 30º (as shown in the figure uploaded). The thread goes through a little hole O in a slab and it is slowly pulled up until the angle Θ becomes 60º. What is the thread's length pulled? What is the variation factor of velocity?

Homework Equations

L = rxmv, T*senΘ=mv²/R and T*cosΘ-mg = 0

The Attempt at a Solution

I've tried solving this question this way:
"let T1 be the tension when θ1=30º, m*g be the weight and v1 be the velocity. Then we can get tg30º = (m*(v1)²/R1)/(mg) = √3/3 => √3/3*g=v²1/R1 (I)
and when Θ=60º we have √3g=v²2/R2. (II)
From these equations I divided I by II:
1/3= (v1/v2)²*R2/R1 and obtained (III). But we threen unknows (v2, R2, v1) and only one equation.

I imagine the angular momentum might conserve in the perpendicular direction to the plane of motion of the mass once it is slowly pulled up, but I have no idea how to apply this.

P.S.: I used R but it is the same in meaning as d in the figure.

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You have two equations, (I) and (II). Add to them conservation of angular momentum, and you will have three equations and three unknowns. Solve.

That's the problem. I don't know how to apply conservation of angular momentum. I tried M*v1*(R1/tgθ1) = M*v2*(R2/tgθ2) where R/tgθ is the distance from the point O to the plane of motion and it was incorrect. Can you tell me why?

You said that ##R## was the same as ##d## in the figure. Then angular momentum is simply ##m v_1 R_1## and ##m v_2 R_2##.

1 person

Hello, great job in attempting to solve this problem. You are correct in thinking that angular momentum is conserved in this system. In fact, angular momentum conservation is a fundamental principle in physics that states that the total angular momentum of a system remains constant unless acted upon by an external torque.

In this case, the system consists of the ball, the thread, and the slab. Initially, the ball is moving in a circular motion with a constant velocity and the thread is at an angle of 30º. When the thread is pulled up, the angle changes to 60º, but the ball continues to move in a circular motion. This means that the angular momentum of the ball remains constant.

Using the equation L = rxmv, where r is the distance from the axis, m is the mass, v is the velocity, and x is the distance from the axis to the center of mass, we can see that the angular momentum is directly proportional to the velocity. This means that as the thread is pulled up, the velocity of the ball must increase in order to keep the angular momentum constant.

To solve for the length of the thread pulled, we can use the equation for centripetal force, T = mv²/R, where T is the tension in the thread, m is the mass, v is the velocity, and R is the radius of the circular motion. We know that the tension in the thread remains constant, so we can set the equations for tension at the two different angles equal to each other and solve for the velocity at each angle. This will give us two equations with two unknowns (v1 and v2).

Once we have the velocities, we can use the equation for angular momentum to solve for the length of the thread pulled. We know that the angular momentum remains constant, so we can set the equations at the two different angles equal to each other and solve for the length of the thread pulled.

As for the variation factor of velocity, we can use the equation v2/v1 = √(R1/R2), where v1 is the initial velocity, v2 is the final velocity, R1 is the initial radius, and R2 is the final radius. This will give us the ratio of the two velocities, which can then be used to find the variation factor.

I hope this helps and good luck with your homework!

1. What is angular momentum conservation?

Angular momentum conservation is a fundamental principle in physics that states that the total angular momentum of a system remains constant unless acted upon by an external torque. In simpler terms, it means that the rotational motion of a system remains the same unless an external force is applied.

2. How is angular momentum conserved?

Angular momentum is conserved because of the law of conservation of angular momentum, which states that for a closed system, the total angular momentum remains constant. This means that the initial angular momentum of a system will be equal to the final angular momentum, even if there are external forces acting on the system.

3. What are some examples of angular momentum conservation?

One common example of angular momentum conservation is seen in ice skaters. When they spin with their arms and legs extended, their angular momentum is low. But when they bring their arms and legs in, their angular momentum increases due to the decrease in their moment of inertia, causing them to spin faster. Another example is seen in the rotation of planets around the sun, which is due to the conservation of angular momentum.

4. How does angular momentum conservation differ from linear momentum conservation?

Angular momentum conservation deals with the rotational motion of a system, while linear momentum conservation deals with the translational motion of a system. Additionally, angular momentum takes into account the distance from the axis of rotation, while linear momentum does not consider distance.

5. What are the practical applications of angular momentum conservation?

Angular momentum conservation has several practical applications, including in the design of vehicles and machinery. For example, the conservation of angular momentum is used in the design of gyroscopes, which are used in airplanes, satellites, and ships to maintain stability and orientation. It is also used in the design of bicycles, where the spinning motion of the wheels helps to maintain balance and stability.

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