Acceleration and Gravity with Circular Motion

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  • #1
Joshua A
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Homework Statement


You hold a small ice cube near the top edge of a hemispherical bowl of radius 100 mm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction.

Homework Equations


ΣF = ma
Fg = mg

The Attempt at a Solution


I know that when the ice cube is held at the top of the bowl (horizontal), there is a downward force due to gravity and no centripetal force. This downward force causes the ice cube to accelerate downwards. The instant the ice cube begins to slide down, there is an increasing centripetal force (equal to the force of gravity that is perpendicular to the bowl) in order to change the direction of the ice cube.

At the bottom of the bowl, the entire force of gravity is perpendicular to the bowl. I know that the ice cube will continue in circular motion so it's direction cannot change and there must be some acceleration at the bottom pointing upwards. This acceleration is due to the normal contact force the bowl exerts on the ice cube.

ΣF = ma
ΣF = Fnormal - Fg = Fnormal - mg = ma

a = Fnormal/m - g

The magnitude of the acceleration given from that equation will give me the answer to the question. However, I do not know how to find out what the normal force is.

Could someone help me figure out how to finish my answer? If there is a better/easier way to approach this problem, let me know.
 
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  • #2
Joshua A said:

Homework Statement


You hold a small ice cube near the top edge of a hemispherical bowl of radius 100 mm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction.

Homework Equations


ΣF = ma
Fg = mg

The Attempt at a Solution


I know that when the ice cube is held at the top of the bowl (horizontal), there is a downward force due to gravity and no centripetal force. This downward force causes the ice cube to accelerate downwards. The instant the ice cube begins to slide down, there is an increasing centripetal force (equal to the force of gravity that is perpendicular to the bowl) in order to change the direction of the ice cube.

At the bottom of the bowl, the entire force of gravity is perpendicular to the bowl. I know that the ice cube will continue in circular motion so it's direction cannot change and there must be some acceleration at the bottom pointing upwards. This acceleration is due to the normal contact force the bowl exerts on the ice cube.

ΣF = ma
ΣF = Fnormal - Fg = Fnormal - mg = ma

a = Fnormal/m - g

The magnitude of the acceleration given from that equation will give me the answer to the question. However, I do not know how to find out what the normal force is.

Could someone help me figure out how to finish my answer? If there is a better/easier way to approach this problem, let me know.
Joshua A said:

Homework Statement


You hold a small ice cube near the top edge of a hemispherical bowl of radius 100 mm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction.

Homework Equations


ΣF = ma
Fg = mg

The Attempt at a Solution


I know that when the ice cube is held at the top of the bowl (horizontal), there is a downward force due to gravity and no centripetal force. This downward force causes the ice cube to accelerate downwards. The instant the ice cube begins to slide down, there is an increasing centripetal force (equal to the force of gravity that is perpendicular to the bowl) in order to change the direction of the ice cube.

At the bottom of the bowl, the entire force of gravity is perpendicular to the bowl. I know that the ice cube will continue in circular motion so it's direction cannot change and there must be some acceleration at the bottom pointing upwards. This acceleration is due to the normal contact force the bowl exerts on the ice cube.

ΣF = ma
ΣF = Fnormal - Fg = Fnormal - mg = ma

a = Fnormal/m - g

The magnitude of the acceleration given from that equation will give me the answer to the question. However, I do not know how to find out what the normal force is.

Could someone help me figure out how to finish my answer? If there is a better/easier way to approach this problem, let me know.
 
  • #3
Take an instance slightly making angle theta with vertical line then resolve the force
 
  • #4
Abhishek kumar said:
Take an instance slightly making angle theta with vertical line then resolve the force

I'm not quite sure what you mean...

If I choose an angle of 5 degrees from the vertical, then the tangential acceleration of the ice cube will be 9.8m/s2 * cos(5) = 9.76m/s2. The component of the acceleration due to gravity perpendicular is 9.8m/s2 * sin(5) = 0.85m/s2 in the direction away from the bowl (so the bowl must exert a normal force that counteracts this as well as changes the direction of the cube). However I'm not sure where to go from here...
 
  • #5
Welcome to PF Joshua ,

Can you find speed of the cube at the bottom of the bowl in terms of the radius of the bowl using one of the conservation laws ?

Can you express centripetal acceleration of cube in terms of its speed at the bottom ?

Does the cube have tangential acceleration at the bottom of the bowl ?
 
  • #6
conscience said:
Welcome to PF Joshua ,

Can you find speed of the cube at the bottom of the bowl in terms of the radius of the bowl using one of the conservation laws ?

Can you express centripetal acceleration of cube in terms of its speed at the bottom ?

Does the cube have tangential acceleration at the bottom of the bowl ?

I can use conservation of energy to find the final velocity:

mgh = 1/2mv2

so v = sqrt(2gh)

I know that centripetal acceleration, ac can be found using:
ac = v2/r

but h and r are the same, so...

ac = 2g = 19.6m/s2

So the centripetal acceleration is 2g. At the bottom of the bowl, there is no tangential acceleration however there is also a downward acceleration due to gravity (-g). Therefore the net acceleration (what I assume the question is asking for) is g.

If the centripetal acceleration is 2g then there must be a normal force equal to 2mg. However, in this thread, the correct answer is that the normal force is 3mg... https://www.physicsforums.com/threads/angular-acceleration-at-the-bottom.836852/
 
  • #7
Joshua A said:
I can use conservation of energy to find the final velocity:

mgh = 1/2mv2

so v = sqrt(2gh)

I know that centripetal acceleration, ac can be found using:
ac = v2/r

but h and r are the same, so...

ac = 2g = 19.6m/s2

So the centripetal acceleration is 2g.

Good .
Joshua A said:
At the bottom of the bowl, there is no tangential acceleration however there is also a downward acceleration due to gravity (-g). Therefore the net acceleration (what I assume the question is asking for) is g

Centripetal acceleration is the net acceleration of the cube towards the center of the bowl .

Writing ∑F = Ma in radial direction .

N -Mg = Mac .

You have already calculated ac =2g .

Putting in the above equations , you get N = 3Mg
 
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  • #8
conscience said:
Centripetal acceleration is the net acceleration of the cube towards the center of the bowl .

I was unaware of this. Does the centripetal acceleration always represent the net acceleration? (Would the net force represent the centripetal force in that case as well?)
 
  • #9
Joshua A said:
Does the centripetal acceleration always represent the net acceleration?

Centripetal acceleration is the component of net acceleration in radial direction (towards the center ) .

But in this problem since there is no tangential acceleration , centripetal acceleration is also the net acceleration .
Joshua A said:
Would the net force represent the centripetal force in that case as well?)

Again , product of mass times centripetal acceleration is centripetal force i.e component of net force towards the center .
 
  • #10
conscience said:
Centripetal acceleration is the component of net acceleration in radial direction (towards the center ) .

But in this problem since there is no tangential acceleration , centripetal acceleration is also the net acceleration .

Again , product of mass times centripetal acceleration is centripetal force i.e component of net force towards the center .

Thank you!
 

1. What is acceleration in circular motion?

In circular motion, acceleration is the rate of change of velocity in a circular path. It is a vector quantity that is directed towards the center of the circle and is constantly changing in magnitude and direction.

2. How is acceleration related to gravity in circular motion?

In circular motion, acceleration is directly proportional to the force of gravity acting on the object. This means that the greater the force of gravity, the greater the acceleration, and vice versa.

3. What is the difference between centripetal and centrifugal acceleration?

Centripetal acceleration refers to the acceleration towards the center of the circle, while centrifugal acceleration refers to the acceleration away from the center of the circle. Centripetal acceleration is caused by an inward force, while centrifugal acceleration is caused by the inertia of the object.

4. How does the radius of the circle affect acceleration in circular motion?

The radius of the circle has a direct effect on the acceleration in circular motion. The smaller the radius, the greater the acceleration, and vice versa. This is because the force of gravity is inversely proportional to the square of the radius, so a smaller radius means a greater force of gravity and therefore a greater acceleration.

5. How does angular velocity affect acceleration in circular motion?

Angular velocity, which is the rate of change of angular displacement, has a direct effect on acceleration in circular motion. The greater the angular velocity, the greater the acceleration, and vice versa. This is because an increase in angular velocity means the object is moving faster along its circular path, resulting in a greater change in velocity and therefore a greater acceleration.

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