Acceleration and Gravity with Circular Motion

[Joshua A]

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
F_g = 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 = F_normal - F_g = F_normal - mg = ma

a = F_normal/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.

 

[Abhishek kumar]

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
F_g = 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 = F_normal - F_g = F_normal - mg = ma

a = F_normal/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.

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
F_g = 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 = F_normal - F_g = F_normal - mg = ma

a = F_normal/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.

 

[Abhishek kumar]

Take an instance slightly making angle theta with vertical line then resolve the force

 

[Joshua A]

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...

 

[conscience]

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 ?

 

[Joshua A]

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, a_c can be found using:
a_c = v²/r

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

a_c = 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/

 

[conscience]

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 .

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 = Ma_c .

You have already calculated a_c =2g .

Putting in the above equations , you get N = 3Mg

 

[Joshua A]

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?)

 

[conscience]

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 .

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 .

 

[Joshua A]

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!