Energy conservation in circular motion

In summary, the conversation discusses a problem involving an object of mass 1 kg moving along a frictionless circular path from point A to point B at an angle of 30 degrees. The question asks for the normal force at point B, with four answer choices provided. The conversation also includes the equations used to solve the problem and a correction made to the initial solution. The correct answer is 25.98 N.
  • #1
Fatima Hasan
319
14

Homework Statement


An object of mass m = 1 kg moves along a circular frictionless path of radius R = 0.5 m from point A to point B at which it makes an angle θ with the vertical direction as shown in the figure. If the object is release from rest (VA = 0) and θ = 30° , what is the normal force at point B (in N) ?
upload_2018-1-6_16-25-52.png


A ) 2.98 B) 25.98 C) 28.19 D) 29.54

Homework Equations


gif.gif


3. The Attempt at a Solution [/B]
Here's my work :
%5CF%3Dma%5C%5CF%3D2.67%20N%5C%5Cmg%20cos%20%5CTheta%20-FN%3D2.67%5C%5C8.6-FN%3D2.67%5C%5CFN%3D5.gif
 

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    %5CF%3Dma%5C%5CF%3D2.67%20N%5C%5Cmg%20cos%20%5CTheta%20-FN%3D2.67%5C%5C8.6-FN%3D2.67%5C%5CFN%3D5.gif
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  • #2
Fatima Hasan said:

Homework Statement


An object of mass m = 1 kg moves along a circular frictionless path of radius R = 0.5 m from point A to point B at which it makes an angle θ with the vertical direction as shown in the figure. If the object is release from rest (VA = 0) and θ = 30° , what is the normal force at point B (in N) ?
View attachment 217967

A ) 2.98 B) 25.98 C) 28.19 D) 29.54

Homework Equations


View attachment 217968

3. The Attempt at a Solution [/B]
Here's my work :
View attachment 217969
Your work is wrong. What is k?
You can not write gh=gh+0.5v^2. It means that the v2=0.
 
  • #3
ehild said:
Your work is wrong. What is k?
You can not write gh=gh+0.5v^2. It means that the v2=0.
K is a typo. It should be m
gh1 = gh2 + 0.5 v^2
 
  • #4
Fatima Hasan said:
K is a typo. It should be m
gh1 = gh2 + 0.5 v^2
It is better now, but your v2is wrong.
 
  • #5
ehild said:
It is better now, but your v2is wrong.
I found the mistake. The height in B, I should substract it from 0.5
The answer is 25.98
 
  • #6
Fatima Hasan said:
I found the mistake. The height in B, I should substract it from 0.5
The answer is 25.98
Well done!
 
  • Like
Likes Fatima Hasan

Related to Energy conservation in circular motion

1. What is energy conservation in circular motion?

Energy conservation in circular motion is the principle that states that the total energy of a system remains constant throughout the motion, even if the speed or direction of the object changes. This means that the sum of kinetic and potential energy remains constant, and no energy is lost or gained during the motion.

2. How does energy conservation apply to circular motion?

In circular motion, the centripetal force acting on the object is always perpendicular to the velocity, which means that the work done by this force is zero. Therefore, the total energy of the object remains constant as it moves in a circular path, as no external work is being done on the object.

3. Can energy be lost in circular motion?

No, according to the principle of energy conservation, energy cannot be lost in circular motion. While the kinetic and potential energy may change, the total energy of the system remains constant.

4. How does the speed affect energy conservation in circular motion?

As the speed of an object in circular motion increases, its kinetic energy also increases. However, this increase in kinetic energy is balanced by a decrease in potential energy, keeping the total energy of the system constant.

5. What is an example of energy conservation in circular motion?

A simple example of energy conservation in circular motion is a rotating object, such as a spinning top or a merry-go-round. As these objects rotate, their kinetic energy increases while their potential energy decreases, keeping the total energy of the system constant.

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