Finding the Velocity of a Mass in Circular Motion

In summary, the problem involves determining the speed of a bob at the top of its path, given the conservation of momentum and the centripetal force equations. The attempt at a solution initially results in a factor of 2, but upon further consideration, it is clear that the motion of the bob involves a variable centripetal force and energy conservation should be used instead.
  • #1
barryj
854
51

Homework Statement


See attached diagram and statement of problem

Homework Equations


Conservation of momentum ...
mv = MV + mv/2
mv/2 = MV
V = mv/(2M)

centripetal force...centripetal force equals weight at top of circle.
MV^2/L = Mg
V^2 = Lg
V = (Lg)^.5

The Attempt at a Solution



Set V = V

mv/(2M) = (Lg)^.5
v = 2M(Lg)^.5/m

The answer is supposed to be v = 4M(Lg)^.5/m
I am off by a factor of 2 where did I go wrong?
 

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  • #2
barryj said:
Set V = V
Won't M slow down as it moves from the bottom to the top?
 
  • #3
1. What is the speed at the top of the path when the bob just barely makes it?
2. Given you answer in 1, what is the centripetal acceleration?
 
  • #4
barryj said:
centripetal force...centripetal force equals weight at top of circle.
MV^2/L = Mg
V^2 = Lg
V = (Lg)^.5
This would be valid if the stiff rod were replaced by a string. As @kuruman noted, you need to reconsider what the required speed of M must be in order to barely make it over the top.
 
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  • #5
I think I see the error. I will fix the error and re post tomorrow.
 
  • #6
barryj said:

Homework Statement


See attached diagram and statement of problem

Homework Equations


Conservation of momentum ...
mv = MV + mv/2
mv/2 = MV
V = mv/(2M)

centripetal force...centripetal force equals weight at top of circle.
MV^2/L = Mg
V^2 = Lg
V = (Lg)^.5

The Attempt at a Solution



Set V = V

mv/(2M) = (Lg)^.5
v = 2M(Lg)^.5/m

The answer is supposed to be v = 4M(Lg)^.5/m
I am off by a factor of 2 where did I go wrong?
 
  • #7
Think, is the motion of the bob with a constant centripetal force or variable centripetal force. In all situations energy has to be conserved so why not try it out?
 
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Related to Finding the Velocity of a Mass in Circular Motion

1. What is circular motion and how is it different from other types of motion?

Circular motion is the movement of an object along a circular path. It is different from other types of motion, such as linear motion, because the direction of the object is constantly changing as it moves along the circular path.

2. What is the velocity of an object in circular motion?

The velocity of an object in circular motion is the speed at which the object is moving along the circular path, as well as the direction in which it is moving. This means that the velocity is constantly changing as the object moves along the circular path.

3. How do you calculate the velocity of an object in circular motion?

The velocity of an object in circular motion can be calculated by dividing the distance traveled along the circular path by the time it took to travel that distance. This can be represented by the equation v = s/t, where v is the velocity, s is the distance, and t is the time.

4. What is centripetal acceleration and how does it relate to circular motion?

Centripetal acceleration is the acceleration that an object experiences as it moves along a circular path. It is always directed towards the center of the circle and is responsible for keeping the object moving along the circular path. In circular motion, the velocity and centripetal acceleration are always perpendicular to each other.

5. Are there any real-world applications of finding the velocity of a mass in circular motion?

Yes, there are many real-world applications of finding the velocity of a mass in circular motion. Some examples include calculating the speed of a satellite in orbit around the Earth, determining the velocity of a car going around a curve, and measuring the speed of a roller coaster on a loop. Understanding the velocity of a mass in circular motion is also crucial in designing and analyzing machines such as centrifuges and amusement park rides.

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