Angular momentum and Springs (2 ques)

In summary, the first problem involves a student sitting on a rotating stool holding two 3.0 kg objects. When his arms are extended, he rotates at 0.75 rad/s. The student then pulls the objects closer to the axis of rotation, causing a change in angular speed. The second problem involves a 15.0-kg object on a frictionless incline with a wheel of radius 11.0 cm. The acceleration of the object is measured to be 2.00 m/s^2. The student is struggling to find the tension of the rope and has tried several incorrect methods.
  • #1
zaxor0
2
0
I have 2 problems that have been giving a bit of trouble since last night when I gave up on them. I found a similar one to the first one on here but I didn't quite understand the equation that was given.

Homework Statement


A student sits on a rotating stool holding two 3.0 kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation, and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg·m2 and is assumed to be constant. The student then pulls the objects horizontally to 0.30 m from the rotation axis.

Find the new angular speed of the student in rad/s.

and...

A 15.0-kg object is attached to a cord that is wrapped around a wheel of radius r = 11.0 cm (Fig. P8.60). The acceleration of the object down the frictionless incline is measured to be 2.00 m/s2. Assume the axle of the wheel to be frictionless. The object is also on a slope of 37 degrees.

(a) Find the tension of the rope.

Homework Equations


for the first one I can not figure out what equation to use at all, a boost in the right direction would help me tremendously. I worked on this last night so I don't remember how exactly I got the following answers but here are the wrong answers I came up with: 1.267 rad/s , 2.757 rad/s , 1.688 rad/s .

for the second, I swear I am going this right but I keep getting the wrong answer.
For the tension of the rope, at first I tried 15kg(9.81m/s^2)cos(37). But the object is accelerating at 2.00m/s^2 so I tried 15kg(2m/s^2) = 30n. Then I tried both added together. These are the three wrong answers I have gotten 117.5 N, 30 N, and 147.5 N.Any help would be greatly appreciated, thank you.

EDIT: didnt mean to put "springs in the title"
 
Last edited:
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  • #2
I solved the first problem, I was using IiWi = IfWf , sorry for lazy latex. But I didnt consider that the given interia of 3 for the student did not needed to be changed. I was halving it using 1/2MR^2 instead of considering as that whole thing.

But as for the other problem I am still stumped.
 
  • #3

For the first problem, you can use the conservation of angular momentum equation, which states that the initial angular momentum is equal to the final angular momentum. In this case, the initial angular momentum is the product of the moment of inertia (3.0 kg·m2) and the initial angular speed (0.75 rad/s). The final angular momentum can be calculated by using the new distance (0.30 m) and the new moment of inertia (3.0 kg·m2 + 2(3.0 kg)(0.30 m)2). This will give you the new angular speed of the student.

For the second problem, you can use the equation for the net force in the y-direction, which is equal to the mass times the acceleration in the y-direction. In this case, the net force is equal to the tension in the rope minus the component of the weight in the y-direction, which is equal to the mass (15.0 kg) times the acceleration (2.00 m/s2) times the sine of the angle (37 degrees). This will give you the tension in the rope.

I would recommend double checking your calculations and making sure that you are using the correct units. It is also helpful to draw a diagram and label all the given values to help you visualize the problem better. Good luck!
 

1. What is angular momentum?

Angular momentum is a measure of the rotation of an object around a fixed axis. It is calculated by multiplying the mass of the object, its velocity, and its distance from the axis of rotation.

2. How is angular momentum related to springs?

Angular momentum is conserved in a system, which means it remains constant unless acted upon by an external force. In the case of a spring, when it is compressed or stretched, the angular momentum of the system remains the same as long as there are no external forces acting on it.

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