Which will reach the bottom faster: solid or hollow sphere?

[Milo Martian]

1. The problem statement:
Consider a solid sphere and a hollow sphere, of equal mass M and equal radius R ,at rest on top of an incline . If there is no slipping which will reach the bottom faster.

2. Homework Equations :
a_cm = F_ext/M (cm= centre of mass)
angular acceleration= torque_ext/ I ( I = moment of inertia)
I_{hollow sphere}= 2MR²/3
I_{solid sphere}=2MR²/5

3. The Attempt at a Solution :
a_cm should be same for both of them as F_ext and M are same, so they should reach at same time. But angular acceleration of solid sphere will be more since I_{solid sphere}< I_{hollow sphere}and torque is same for both, so by this info the solid sphere should reach faster. Thus, the confusion.Also, how can a_cm remain same for both when angular acceleration of solid sphere is higher , the a_cm of the solid cylinder should be higher since the radius is same for both. But according to the first equation, a_cm should be same.

 

[drvrm]

acm should be same for both of them as Fext and M are same, so they should reach at same time. But angular acceleration of solid sphere will be more since Isolid sphere< Ihollow sphereand torque is same for both, so by this info the solid sphere should reach faster.

how the angular acceleration is related with the acceleration of the center of mass of the two spheres ?
i doubt that a(c.m.) is same for both as the rotation and translational motion seems to be related(a case of pure rolling).

 

[Milo Martian]

how the angular acceleration is related with the acceleration of the center of mass of the two spheres ?
i doubt that a(c.m.) is same for both as the rotation and translational motion seems to be related(a case of pure rolling).

well , if there is no slipping the distance covered in one rotation is 2piR for both the spheres. So, with higher angular acceleration, the number of rotations per second increases at higher rate, leading to higher increase in rate of ground coverage, so shouldn't the a_cm be higher as well? But, is the first equation not applicable for pure rolling?

 

[Vibhor]

a_cm should be same for both of them as F_ext and M are same, so they should reach at same time.

No . The net force is not same . Please write equation of motion for both translation and rotational motion .Find expression for a_cm in terms of given variables .

 

[Milo Martian]

No . The net force is not same . Please write equation of motion for both translation and rotational motion .Find expression for a_cm in terms of given variables .

aren't the equations given the right one's for translation and rotation? isn't the equation , a_cm = Net external force/ Mass ,correct for translation + rotation?

 

[haruspex]

sn't the equation , acm = Net external force/ Mass ,correct for translation + rotation?

Yes, but as Vibhor wrote the net force is not the same. What forces act on the spheres?

 

[Milo Martian]

Yes, but as Vibhor wrote the net force is not the same. What forces act on the spheres?

Well... there's a force mgsin(theta) acting parallel to the incline along the centre of mass and friction acting at the bottom of the sphere in the opposite direction. Since there's no slipping, the forces should be equal.So, the net external force for both is 0? so, a_cm is 0? that can't be right. I'm confused.

 

[haruspex]

mgsin(theta) acting parallel to the incline along the centre of mass

Well, that's the component of the gravitational force in that direction, yes.

Since there's no slipping, the forces should be equal.

No such rule.

 

[Milo Martian]

Well, that's the component of the gravitational force in that direction, yes.

No such rule.

But as the body does not slip , shouldn't the magnitude of static friction be equal to the magnitude of the external force trying to make it move?

 

[haruspex]

But as the body does not slip , shouldn't the magnitude of static friction be equal to the magnitude of the external force trying to make it move?

The fact of non-slipping only tells you that the point of the wheel in contact with the plane has no acceleration or velocity parallel to the plane. The net zero acceleration will be achieved by the linear downslope acceleration of the wheel as a whole being equal and opposite to the tangential acceleration resulting from the wheel's rotational acceleration.

 

[Milo Martian]

The fact of non-slipping only tells you that the point of the wheel in contact with the plane has no acceleration or velocity parallel to the plane. The net zero acceleration will be achieved by the linear downslope acceleration of the wheel as a whole being equal and opposite to the tangential acceleration resulting from the wheel's rotational acceleration.

Yes, i thought i said that only. The linear downslope acceleration comes from the component of gravitational force in that direction ( doesn't it) and the tangential acceleration comes from friction (doesn't it?) Since force= mass X acceleration and the mass and acceleration being equal , the forces will also be equal ( isn't it?).
I am getting really confused, the answer to my doubts seems miles away. please help

 

[haruspex]

Yes, i thought i said that only. The linear downslope acceleration comes from the component of gravitational force in that direction ( doesn't it) and the tangential acceleration comes from friction (doesn't it?) Since force= mass X acceleration and the mass and acceleration being equal , the forces will also be equal ( isn't it?).
I am getting really confused, the answer to my doubts seems miles away. please help

The wheel is a rigid body. You cannot figure out the motion of one small piece of it just from looking at external forces applied to the wheel as a whole. There are also forces between the parts of the wheel.
The two accelerations only cancel out at the point of contact.

The normal force and the component of the gravitational force perpendicular to the plane exactly cancel and are in the same line of action, so they produce no net force nor torque. So we can ignore these.

There is the downslope component of gravity, call this force F_gd, and the upslope force of friction, F_f. The difference between them gives the net force which provides the linear downslope acceleration of the wheel.

At the same time, these two forces produce a net torque. That produces the rotational acceleration.
The value of the torque depends where you put your reference axis. If we put it at the centre of the wheel then it will be R.F_f. If we put it at the point of contact it will be R.F_gd. The corresponding moments of inertia for those two axes are also different, and it turns out that the resulting angular acceleration is the same either way.

 

[Milo Martian]

The wheel is a rigid body. You cannot figure out the motion of one small piece of it just from looking at external forces applied to the wheel as a whole. There are also forces between the parts of the wheel.
The two accelerations only cancel out at the point of contact.

The normal force and the component of the gravitational force perpendicular to the plane exactly cancel and are in the same line of action, so they produce no net force nor torque. So we can ignore these.

There is the downslope component of gravity, call this force F_gd, and the upslope force of friction, F_f. The difference between them gives the net force which provides the linear downslope acceleration of the wheel.

At the same time, these two forces produce a net torque. That produces the rotational acceleration.
The value of the torque depends where you put your reference axis. If we put it at the centre of the wheel then it will be R.F_f. If we put it at the point of contact it will be R.F_gd. The corresponding moments of inertia for those two axes are also different, and it turns out that the resulting angular acceleration is the same either way.

Ok.. so to find out a_cm , need to find out angular acceleration and multiply it by R?
and regarding the equation a_cm= F_ext/M , can't we predict the motion of centre of mass from the knowledge of external forces and the mass? but the forces due to different parts of the sphere are internal forces and are not part of the above equation. Sorry, if my questions are getting repetitive .

 

[Nidum]

It is always easier to understand these problems using a diagram . Take a copy of the diagram above and mark in whatever forces that you think are acting

 

[Milo Martian]

View attachment 104317
It is always easier to understand these problems using a diagram . Take a copy of the diagram above and mark in whatever forces that you think are acting

OK so,m being mass of the body, here normal force, N = mgcosA and since there is no slipping ( so there is only static friction) the static friction, f should be equal to mgsinA i think. There will be a torque due to the combination of friction, f and component of gravitational force mgsinA. Am i wrong so far? If yes, can you tell me where and why?
If i am not wrong then are the equations i gave in the question applicable here? if yes, then since friction, f and mgsinA are equal and opposite so the net force on the body is 0 right? So, a_cm= 0 ( since F_ext=0) which must not be right. Or is F_ext= mgsinA as friction is internal force to the system and, thus, the a_cm= gsinA? Here is my confusion. Can you tell me how to get the right a_cm? and what's wrong with my method?

 

[haruspex]

since there is no slipping ( so there is only static friction) the static friction, f should be equal to mgsinA

You keep repeating that, but it is not true. Your intuition is blocking you. Just work with the free body diagram and the standard laws, F_net=ma, τ_net=Iα, and the rolling condition a=rα.
Fot the torque equation, make sure to use the same axis for both torque and moment of inertia.
When you have worked that through, you can see for yourself how the static frictional force differs from mg sin θ.

 

[Milo Martian]

You keep repeating that, but it is not true. Your intuition is blocking you. Just work with the free body diagram and the standard laws, F_net=ma, τ_net=Iα, and the rolling condition a=rα.
Fot the torque equation, make sure to use the same axis for both torque and moment of inertia.
When you have worked that through, you can see for yourself how the static frictional force differs from mg sin θ.

Ok so f not equal to mgsinA. But F_ext= mgsinA, right? so a_cm = gsinA for both the spheres?

 

[haruspex]

But Fext= mgsinA, right?

What is F_ext? If you mean F_net, no.

a_cm= gsinA for both the spheres?

No.

Quit guessing and follow proper procedure:

• There are two forces parallel to the plane, mg sin A and F_friction
• What is the net force?
• Write the equation for linear acceleration
• What is the torque about the mass centre?
• Write the equation for angular acceleration
• Because it is rolling contact, the angular acceleration and linear acceleration are related. Write the equation for that.

 

[Milo Martian]

What is F_ext? If you mean F_net, no.

No.

Quit guessing and follow proper procedure:

• There are two forces parallel to the plane, mg sin A and F_friction
• What is the net force?
• Write the equation for linear acceleration
• What is the torque about the mass centre?
• Write the equation for angular acceleration
• Because it is rolling contact, the angular acceleration and linear acceleration are related. Write the equation for that.

F_ext= net force external to the system of incline+ sphere. isn't friction internal to the system?if yes then the only external force is mgsinA, isn't it?

 

[Milo Martian]

What is F_ext? If you mean F_net, no.

No.

Quit guessing and follow proper procedure:

• There are two forces parallel to the plane, mg sin A and F_friction
• What is the net force?
• Write the equation for linear acceleration
• What is the torque about the mass centre?
• Write the equation for angular acceleration
• Because it is rolling contact, the angular acceleration and linear acceleration are related. Write the equation for that.

What is F_ext? If you mean F_net, no.

Oh i got my mistake. F_ext= force external to the sphere = mgsinA-f. right?

 

[Milo Martian]

What is F_ext? If you mean F_net, no.

No.

Quit guessing and follow proper procedure:

• There are two forces parallel to the plane, mg sin A and F_friction
• What is the net force?
• Write the equation for linear acceleration
• What is the torque about the mass centre?
• Write the equation for angular acceleration
• Because it is rolling contact, the angular acceleration and linear acceleration are related. Write the equation for that.

ok, so F_ext= mgsinA - f = ma_cm
=> a_cm= (mgsinA-f)/m ...(1)

Torque= fXR ( taken about cm)= I X angular acceleration
angular acc. for solid sphere= fXR/ I_{solid sphere} = 5fXR/ 2mR²=5f/2mR
linear acc. of cm for solid sphere= (5f/2mR) X R= 5f/2m ...(2)
angular acc. for hollow sphere= fXR/ I_{hollow sphere}= 3f/2mR
linear acc. of cm for hollow sphere= 3f/2m ...(3)
From equation (1) i am getting same a_cm for both.
From equation (2) and (3) i am getting diff a_cm for solid and hollow sphere.
where am i going wrong? the only assumption i made is that f is same for both the spheres. is that assumption wrong?

 

[haruspex]

the only assumption i made is that f is same for both the spheres. is that assumption wrong?

Yes, it is wrong. You do not need to make any assumptions.

 

[Milo Martian]

Yes, it is wrong. You do not need to make any assumptions.

ok, so rest of my calculation is right?
let the two fs be f1 and f2.
a_cm for solid sphere=5f1/2m
a_cm for hollow sphere= 3f2/2m
so how do i figure out which a_cm is more , since i don't know which among f1 and f2 is greater, and thus solve the initial question of which sphere reaches the bottom first?

 

[llober]

For the same mass, the hollow sphere has a greater "moment of inertia", so will opose more resistance to rotational angular acceleration... seems to me that hollow sphere will be slower

 

[Milo Martian]

For the same mass, the hollow sphere has a greater "moment of inertia", so will opose more resistance to rotational angular acceleration... seems to me that hollow sphere will be slower

can u look at my calculations and tell me where i am going wrong?

 

[haruspex]

et the two fs be f1 and f2.
acm for solid sphere=5f1/2m

It depends what you mean by the "f"s. You were earlier using f for the frictional force. For that equation you need the net force.

so how do i figure out which a_cm is more , since i don't know which among f1 and f2 is greater, and thus solve the initial question of which sphere reaches the bottom first?

You need to do the other steps in post #20.

 

[llober]

can u look at my calculations and tell me where i am going wrong?

I think is better to do an Energy balance...
https://www.physicsforums.com/attachments/104346

 

[haruspex]

I think is better to do an Energy balance...

Please delete ths post. Violates protocol. we are not supposed to lay out complete solutions, just provide hints and corrections. please read the forum guidelines.
Anyway, it is erroneous.

 

[Milo Martian]

It depends what you mean by the "f"s. You were earlier using f for the frictional force. For that equation you need the net force.

You need to do the other steps in post #20.

Okay, i got it .
F_ext= mgsinA - f= ma_cm ...(1)
fXR= torque = I X angular acc. = I X a_cm/R
=> f= Ia_cm/R² ...(2)
From (1) and (2) :
ma_cm = mgsinA - Ia_cm/R²
=> a_cm= mgsinA/(m + I/R²)
So, since I_{solid sphere}< I_{hollow sphere}
Therefore, a_cm for solid sphere> a_cm for hollow sphere.
i hope i am right. phew

 

[Milo Martian]

I think is better to do an Energy balance...
https://www.physicsforums.com/attachments/104346

yes, that can also work for solving the question. But my main confusion was regarding finding the a_cm. But thanks for your help.

 

[Milo Martian]

I think is better to do an Energy balance...
https://www.physicsforums.com/attachments/104346

But won't mechanical energy change due to friction? just curious

 

[llober]

But won't mechanical energy change due to friction? just curious

If there's no sliding, there's no loss by friction, because the no sliding translates into speed zero at the contact point, so the "friction force" does not move.

 

[Milo Martian]

If there's no sliding, there's no loss by friction, because the no sliding translates into speed zero at the contact point, so the "friction force" does not move.

OK

 

[haruspex]

Okay, i got it .
F_ext= mgsinA - f= ma_cm ...(1)
fXR= torque = I X angular acc. = I X a_cm/R
=> f= Ia_cm/R² ...(2)
From (1) and (2) :
ma_cm = mgsinA - Ia_cm/R²
=> a_cm= mgsinA/(m + I/R²)
So, since I_{solid sphere}< I_{hollow sphere}
Therefore, a_cm for solid sphere> a_cm for hollow sphere.
i hope i am right. phew

You got it.