Circular motion and moment of inertia - circle loop

[CRJ]

Homework Statement

A small solid porcelain sphere, with a mass m and radius r, is placed on the inclined section of the metal track shown below, such that its lowest point is at a height h above the bottom of the loop. The sphere is then released from rest, and it rolls on the track without slipping. In your analysis, use the approximation that the radius r of the sphere is much smaller than both the radius R of the loop and the height h. (Use the following as necessary: M, R, and g for the acceleration of gravity.)
Find net vertical force at point P

Relevant Equations

I (sphere)= 2/5mr^2
torque= moment of inertia x angular acceleration

 

[PeroK]

It's quite hard to figure out what you're doing when you simply post numbers and equations. Please explain your approach.

 

[nasu]

You seem to assume that the "a" in the Newton's law for the angular acceleration is the centripetal acceleration. Do you think that this is the full acceleration of the Com at that point?

 

[Lnewqban]

Welcome, @CRJ !

Have you tried following the directions shown in red fonts?
Could you explain the 20/7 factor in the value of Fx?

 

[CRJ]

Fx is the horizontal component, which is the centripetal acceleration, so it is mv^2/R. From the previous part as the sphere rolls without slipping the rotational kinetic energy is considered, and v^2= 20/7 gR when the sphere rolls from 3R to R. substituting it into Fc=mv^2/R gives the horizontal (centripetal) force of 20/7 mg.

For vertical I tried -mg too as weight should be the only vertical force acting on it but it is wrong.

So I take torque which is equal to F×R and also equal to I×α (α being angular acceleration). and since it is rolling without slipping a=r×α. Then I took a=v^2/r from circular motion (This one I don't know if it can actually be used but it is the only equation I can think of that links them) Then for sphere, I=(2/5)×mr^2. Combining the equation gives me F=8/7mg. I then minus off the weight as I thought weight would be opposite direction. But now to think of it, it is net torque so maybe I shouldn't minus weight off as it is already Fnet?

I am left with one last try for the question.

 

[CRJ]

You seem to assume that the "a" in the Newton's law for the angular acceleration is the centripetal acceleration. Do you think that this is the full acceleration of the Com at that point?

thank you for your reply.
do you mean torque=I ×α ? I took a=r×α for rolling without slipping and sub it in. do you mean that a found that way is centripetal and not tangential so I can't use it for torque?
I am not really sure also, if it cant be use then how do I find angular acceleration or tangential acceleration? Could you explain a little more?

 

[CRJ]

Welcome, @CRJ !

Have you tried following the directions shown in red fonts?
Could you explain the 20/7 factor in the value of Fx?

Thank you for your reply
I am not really sure about what the directions in red fonts mean by using newtons second law, but I did use τ=Iα for rigid body and the relation for rolling without slipping a_CM= r×α.

 

[haruspex]

I do not understand any of the working for finding Fy. The first line seems to confuse R with r. Please explain the thinking.

For vertical I tried -mg too as weight should be the only vertical force acting on it but it is wrong.

It is not slipping and it is losing speed. Therefore there is vertical friction to reduce its angular speed.
Draw an FBD of the sphere in that position.

 

[CRJ]

I do not understand any of the working for finding Fy. The first line seems to confuse R with r. Please explain the thinking.

It is not slipping and it is losing speed. Therefore there is vertical friction to reduce its angular speed.
Draw an FBD of the sphere in that position.

Thank you for your reply
True, I am not quite sure of which r to use actually.
So does it mean that vertical force should be negative since it is slowing down?
I am not quite sure for FBD, is there a component of the normal force acting upwards? Shouldn't normal force act perpendicular to the surface which makes it Fy? If it is rolling without slipping so there is friction, so does the friction act upwards (like sole of the shoes) or downwards (oppose motion upwards?). Then there is weight -mg acting vertically downwards. But the answer is not -mg so there should actually be another component acting upwards or downwards.

 

[PeroK]

Thank you for your reply
True, I am not quite sure of which r to use actually.
So does it mean that vertical force should be negative since it is slowing down?
I am not quite sure for FBD, is there a component of the normal force acting upwards? Shouldn't normal force act perpendicular to the surface which makes it Fy? If it is rolling without slipping so there is friction, so does the friction act upwards (like sole of the shoes) or downwards (oppose motion upwards?). Then there is weight -mg acting vertically downwards. But the answer is not -mg so there should actually be another component acting upwards or downwards.

Have you studied a wheel or sphere accelerating and decelerating as it rolls without slipping? If not, you should study this, and especially the role of static friction (not kinetic friction).

 

[CRJ]

Have you studied a wheel or sphere accelerating and decelerating as it rolls without slipping? If not, you should study this, and especially the role of static friction (not kinetic friction).

Thank you for your reply
I think I do know a bit as in the relationship between angular and linear velocity, acceleration. I also know that there has to be static friction (not kinetic friction as it is not slipping) as the sphere moves, and it is there but there is no work done by friction as the point on the sphere kind of moves. Attached below are my notes from summary lecture, pretty much what I have learnt up till this point. Will be glad if you could refer me some other good online resources. (I personally watch some physics (&math) youtube videos when I get stuck and need some help but I cant seem to find relevant resources for this part of the question)

 

[PeroK]

Consider a sphere rolling without slipping up an incline. There is the component of the force of gravity acting through the centre of mass down the slope. As the sphere is rolling, the deceleration is not simply given by $g\sin \theta$. Because as the sphere slows down its rotation must also slow down at the appropriate rate. The force that slows the rotation is static friction acting up the slope. It must act up the slope in order to provide a torque against the direction of rotation. But, that force also accelerates the sphere up the slope (or, at least, reduces the deceleration). The force of static friction, therefore, is such that it balances the linear deceleration and the rotational deceleration - and can thus be calculated.

The other side of this phenomena is that objects will roll down a slope with different accelerations depending on their moment of inertia. There's a video demonstration here (start watch at 4:50 if you want to look at objects rolling).

 

[CRJ]

Thank you so much for your help, it is true that I previously didn't consider the friction force (upwards)

So I actually got an answer from Chegg and turns out it is correct (got a classmate to try it)

But I dont fully understand the solution. So there is an upward friction force to be considered. What I dont understand from the above solution is how they take net torque as f (by my understanding f should be representing friction). But why is it that only friction contributes to the net torque? Doesn't weight contribute to net torque? I am quite confused about their use of r and R as well.

 

[PeroK]

Thank you so much for your help, it is true that I previously didn't consider the friction force (upwards)

So I actually got an answer from Chegg and turns out it is correct (got a classmate to try it)
View attachment 352219
But I dont fully understand the solution. So there is an upward friction force to be considered. What I dont understand from the above solution is how they take net torque as f (by my understanding f should be representing friction). But why is it that only friction contributes to the net torque? Doesn't weight contribute to net torque? I am quite confused about their use of r and R as well.

An object's weight acts through its centre of mass and exerts no torque on the object.

 

[CRJ]

Is this FBD correct?

 

[CRJ]

An object's weight acts through its centre of mass and exerts no torque on the object.

Ohh
I previously thought that torque is about the centre of the loop, I thought its the motion around the loop.
So it is actually torque about the centre of mass and accounts for the rotation of the object as it roll up the loop? so small r should be used for the whole equation, and there is no big r. I is moment of inertia of the sphere so it should be using small r, and a=α×r is also about the sphere itself.
thank you very much

 

[PeroK]

View attachment 352224
Is this FBD correct?

Yes.

 

[PeroK]

Ohh
I previously thought that torque is about the centre of the loop, I thought its the motion around the loop.

You can calculate the torque about any point you like. Technically, you should always specify about which point or axis the torque applies.

So it is actually torque about the centre of mass and accounts for the rotation of the object as it roll up the loop?

Yes.

so small r should be used for the whole equation, and there is no big r. I is moment of inertia of the sphere so it should be using small r, and a=α×r is also about the sphere itself.
thank you very much

Yes, it's all about the forces acting directly on the sphere.

 

[haruspex]

So it is actually torque about the centre of mass and accounts for the rotation of the object as it roll up the loop?

The torque on the sphere (about whatever axis you choose) accounts for the angular acceleration of the sphere.
Three forces act on the sphere: the normal force, the frictional force and gravity.
There are two natural ways to approach this:

1. Moments about sphere's centre. The normal force and gravity exert no torque.
2. Moments about point of contact. The normal force and friction exert no torque. Use the parallel axis theorem.

 

[CRJ]

Erm I a bit confused about the sign, like the ans given by Chegg would suggest the solution above. I don't quite get it, they probably do this so that the direction are the same.

but if I actually change it to this, it would give me the same answer too. Would this be better? Like the force for the torque is upwards the direction of the velocity at the point of contact should be downwards and since they are opposite in direction the angular acceleration should be negative? and it makes sense since the sphere probably will slow down?

 

[CRJ]

The torque on the sphere (about whatever axis you choose) accounts for the angular acceleration of the sphere.
Three forces act on the sphere: the normal force, the frictional force and gravity.
There are two natural ways to approach this:

1. Moments about sphere's centre. The normal force and gravity exert no torque.
2. Moments about point of contact. The normal force and friction exert no torque. Use the parallel axis theorem.

I think I understand what you are saying, and I can get the same answer. Is this correct?
(I think angular velocity is clockwise and mg will cause an anticlockwise moment, so angular acceleration will be negative)

 

[haruspex]

View attachment 352236
I think I understand what you are saying, and I can get the same answer. Is this correct?
(I think angular velocity is clockwise and mg will cause an anticlockwise moment, so angular acceleration will be negative)

That all looks fine.
Wrt the sign, what matters is the sign of the answer. Since up is positive, the minus sign is clearly correct. Getting it right through the algebra is tricky. E.g. what is the sign of r in $a=r\alpha$?

 

[CRJ]

That all looks fine.
Wrt the sign, what matters is the sign of the answer. Since up is positive, the minus sign is clearly correct. Getting it right through the algebra is tricky. E.g. what is the sign of r in $a=r\alpha$?

hmm. r should be positive since it is the radius and cannot be negative, and r is a constant since it is the radius of the sphere. a and α should be negative since the sphere decelerates (slows down). so it would be a=r×α or -a=r×(-α)?

I am concerned because if it is done the other way(taking center of mass as pivot), then there will be substitution and the direction or signs have to agree or it will not add up, and it will give a different answer

Sorry I see now that I have assumed negative acceleration to be decelerating, but actually that is wrong, and acceleration is downwards that's why it is negative, and the solution that Chegg gave should be correct. The friction in this case is upwards so the angular acceleration should be positive sign? I think I am still missing something

 

[CRJ]

So net acceleration should be negative since net force is negative and angular acceleration is due to friction force which is positive so it should be positive? So is the angular velocity negative thats why the angular velocity and acceleration have different signs, they are opposite so the sphere decelerates. The velocity at the point of contact should be downwards, I view it as rotating clockwise. But then about the relationship between acceleration and angular velocity, does a=r×α not always hold?

 

[haruspex]

does a=r×α not always hold?

In vectors, sure: $\vec a=\vec r\times\vec \alpha$, where $\vec r$ is the vector from A to B and $\vec a$ is the acceleration at B relative to A.
But you appeared to be using scalars, where it can be confusing. Taking the centre as (0,0), up, right and anticlockwise positive, a point at (r,0) has $a=r\alpha$, while a point at (-r,0) has $a=-r\alpha$. (Compare with the second line of algebra in your post #23.)
Good so far, but what about (0,r)?

Edit:
There is a trick available that allows scalars to be used yet have the signs fall out correctly. See quaternions.
For 2D problems, you can use a simplified version that involves complex numbers. With the usual up, right, anticlockwise positive convention, represent the displacement as $x+iy$ and the angular acceleration (or velocity) as $i\alpha$. $a=(x+iy)i\alpha=(-y+ix)\alpha$, i.e. a horizontal acceleration of $-y\alpha$ and a vertical acceleration of $x\alpha$.

Compare with $(x, y, 0)\times (0, 0, -\alpha)=(-y\alpha, x\alpha, 0)$.

 

[CRJ]

Sorry for the late reply, I was quite busy doing lectures and tutorials for the past few days and didn't really have time. I asked my physics lecturer and he said that ω&α are vectors but they are complicated so it is not yet covered.

So for cross multiplication the direction of the vector is given by the thumb when first vector is the index finger and third finger is the middle finger right?

So for vectors cross multiplication, by right hand rule is it so that z=x×y so z will point out of the paper. Is it by another right hand rule since the thumb point out of the paper anticlockwise is positive. So ω being clockwise it would be negative. Then for α to know the direction of α we have to first find the torque which is r cross F, so for the centre of mass as pivot it would be r to the right F upwards so torque points out of the paper and is positive. torque = I dot α so α is in the same direction as torque so it is positive? So ω is negative and α is positive so the sphere decelerates.

Then by right hand rule a = r cross α ( is there really such a thing? I am not really sure as I previously actually use × as times actually) But if it is true r to the right and α out of paper would give thumbs pointing downwards in the negative y direction so a is negative.

The vector r does it point from the pivot to the point of interest. I did the above with this assumption

 

[CRJ]

Erm if I take r as pointing out radially from center of the sphere then at (r,0) r to the right(x), α positive pointing out of the paper (z) would give me a downwards (y) hence negative. At (-r,0) r to the left(x), α positive pointing out of the paper (z) would give me a upwards (y) hence positive.

then at (0,r), r points up (y) and α points out of paper (z) will give me a to the right (x) hence positive

 

[CRJ]

If I take r as pointing inwards then it would give me the same result as yours for (r,0) and (-r,0), and for (0,r) r is downwards(y), α out of the paper (z) and a would be to the left (x) and hence negative

 

[CRJ]

But α is caused by one force and a is the result of two forces, so is there a relationship between their direction? Or is it just related by the sphere itself (by the circumference etc because the ball roll without slipping) so a is r multiply by α and not cross product of r and α?

 

[CRJ]

Hmm for the quarternion I don't quite get what is got on, will take a look at it (watch some youtube videos) later. Thank you