Ball roll down from the top of a rough spherical dome

[Yoonique]

Homework Statement

A ball is rolling down from the top of a rough spherical dome with negligible initial velocity and angular velocity. Show that the ball must slide before losing the contact with the dome.

Homework Equations

ΣF=ma
Στ = Fr = Iα
f_s = μ_sN
v_cm = rω
Δmgh = 0.5mv_cm² + 0.5I_cmω²
I = 2(mr²)/5

The Attempt at a Solution

Along the sphere:
ΣF = ma
mgsinθ - f_s = ma_tan

Along radius of the sphere:
mgcosθ - N = ma_rad

Στ = Fr = Iα
f_sr = Iα

Let θ₁ be where the ball lose contact. R be the radius of the sphere, r be the radius of the ball.
mgh = 0.5mv_cm² + 0.5I_cmω²
mg(R+r)(1-cosθ₁) = 0.7mv_cm²

When N=0,
mgcosθ₁ = ma_rad
gcosθ₁ = v_cm²/(R+r)
v_cm² = (R+r)gcosθ₁
mg(R+r)(1-cosθ₁) = 0.7(R+r)mgcosθ₁
cosθ₁ = 1/1.7
θ₁ = 54.0°

Let θ₂ be where the ball starts to slide.
To show that the ball slides before it lose contact, θ₂ < θ₁
When is starts to slide, f_s = μ_sN
f_sr = Iα
μ_sN = 2mrα/5
N = 2mrα/5μ_s

And if I sub it into the equations, it gets pretty complicated. I can't solve for θ. So how do I continue from here?

 

[Merlin3189]

Assuming you've worked out when the ball loses contact, when the required centripetal force is greater than the normal (radial) component of weight, the angle where it slides is simply where θ₂ excedes the angle of friction,
tan θ₂ > μ_s, or in other words,
the radial component of weight x μ_s < the tangential component of weight.

 

[Yoonique]

Assuming you've worked out when the ball loses contact, when the required centripetal force is greater than the normal (radial) component of weight, the angle where it slides is simply where θ₂ excedes the angle of friction,
tan θ₂ > μ_s, or in other words,
the radial component of weight x μ_s < the tangential component of weight.

I have already worked out when the ball loses contact which is 54.0°
mgsinθ₂ > μ_sN
sinθ₂ > 2rα/5g
But I got a variable α and I can't solve the value for θ.

 

[Yoonique]

Can anyone give me some hints to solving this question?

 

[Yoonique]

BUMP, went to the second page. Anyone?

 

[PeroK]

BUMP, went to the second page. Anyone?

Just an idea. If you have $\theta_1$ where the ball would fall off, then plug that value back into the equation for rolling without slipping and see whether it balances. Or, whether the frictional force would have to be too large by that stage.

 

[ehild]

When N =0 fs must be also zero, so fs=μN holds. That is slipping!

From energy considerations you get the linear speed of the CM. From that, you get its linear and angular acceleration, and you can obtain expressions for N and for fs as function of the angle theta. Find the condition for theta when fs=μN, and check if it is greater or less than the angle of separation if pure rolling would happen.

 

[PeroK]

When N =0 fs must be also zero, so fs=μN holds. That is slipping!

From energy considerations you get the linear speed of the CM. From that, you get its linear and angular acceleration, and you can obtain expressions for N and for fs as function of the angle theta. Find the condition for theta when fs=μN, and check if it is greater or less than the angle of separation if pure rolling would happen.

Actually, with this insight, you may not have to do any calculations at all!

 

[Yoonique]

If I just plug in the value of θ₁ into the equation, I would get N=0 and f_s = μ(0) = μN (condition for slipping). But this only shows that the moment it loses contact it slips. The question wants me to show that it is already slipping BEFORE it loses contact. So I need to find a value for θ₂ and show that it is less than θ₁.

Okay I will try ehild's method.

 

[PeroK]

If I just plug in the value of θ₁ into the equation, I would get N=0 and f_s = μ(0) = μN (condition for slipping). But this only shows that the moment it loses contact it slips. The question wants me to show that it is already slipping BEFORE it loses contact. So I need to find a value for θ₂ and show that it is less than θ₁.

Okay I will try ehild's method.

Yes, using ehild's insight is the way to go.

For future reference, if you show that $q_1 < q_2$ at some point, then $q_1$ must have been less than $q_2$ before that point. Where $q_1$ and $q_2$ are any continuously changing quantities.

 

[Yoonique]

I tried but my equation simplified to θ₂ and μ_s. μ_s is an unknown, thus I can't solve for θ₂. Is there an equation I missed out?

 

[PeroK]

I tried but my equation simplified to θ₂ and μ_s. μ_s is an unknown, thus I can't solve for θ₂. Is there an equation I missed out?

I would try to solve this problem without any calculations. Sometimes, when the equations get complicated, it's good to solve the problem using logical arguments to minimise the calculations you need.

Here are the two key things you know:

1) The ball starts to roll and gets faster (its energy must keep increasing).

2) The normal force must reduce to 0 by the time it leaves the sphere.

Try to solve it logically using those facts and what you know about the conditions for rolling without slipping.

 

[ehild]

Why don't you show your work in detail?

 

[Yoonique]

Why don't you show your work in detail?

Ok, here it is:

Let θ₂ be where the ball starts to slide.
To show that the ball slides before it lose contact, θ₂ < θ₁

mg(R+r)(1-cosθ₁) = 0.7mv_cm²
v_cm² = (10/7)g(1-cosθ)(R+r)

mgcosθ₂ - N = 10/7 mg(1-cosθ)
N = mg(10/7 cosθ₂ - 10/7)

f_s = 2/3mr²(a/r²) = 2/3 ma

mgsinθ₂ - f_s = ma
mgsinθ₂ = 3.75f_s

When is starts to slide, f_s = μ_sN
mgsinθ₂ = 3.75μ_sN = 3.75μ_smg(10/7 cosθ₂ - 10/7)
sinθ₂ = 3.75μ_s(10/7 cosθ₂ - 10/7)

I can't solve for θ₂ with μ_s being unknown.

 

[Yoonique]

I would try to solve this problem without any calculations. Sometimes, when the equations get complicated, it's good to solve the problem using logical arguments to minimise the calculations you need.

Here are the two key things you know:

1) The ball starts to roll and gets faster (its energy must keep increasing).

2) The normal force must reduce to 0 by the time it leaves the sphere.

Try to solve it logically using those facts and what you know about the conditions for rolling without slipping.

Edited my post after thinking through.

As the ball speeds up, N decreases so f_s decreases thus the tangential acceleration of the ball increases. Since f_s = 2/3ma, f_s increases? But i already said f_s decrease at the start.. I'm confused..

 

[ehild]

When is starts to slide, f_s = μ_sN
mgsinθ₂ = 3.75μ_sN = 3.75μ_smg(10/7 cosθ₂ - 10/7)
sinθ₂ = 3.75μ_s(10/7 cosθ₂ - 10/7)

I can't solve for θ₂ with μ_s being unknown.

You obtained that in case the ball rolled without slipping all the way along the dome, it would leave it at the angle where 10/7 cosθ₂ - 10/7=0. As the left-hand side must be positive, the right-hand side is also positive. This expression decreases with increasing angle. Is the angle where slipping occurs less than the one it would leave the dome if it rolled all the way ?

 

[PeroK]

Edited my post after thinking through.
As the ball speeds up, N decreases so f_s decreases thus the tangential acceleration of the ball increases. Since f_s = 2/3ma, f_s increases? But i already said f_s decrease at the start.. I'm confused..

Suppose the ball rolls without slipping. The tangential force (gravity) is increasing, so the friction force must be increasing. It can't go on increasing if, eventualy, it must get to 0.

 

[Yoonique]

You obtained that in case the ball rolled without slipping all the way along the dome, it would leave it at the angle where 10/7 cosθ₂ - 10/7=0. As the left-hand side must be positive, the right-hand side is also positive. This expression decreases with increasing angle. Is the angle where slipping occurs less than the one it would leave the dome if it rolled all the way ?

Ops I found a mistake in my working, it should be N = mg(17/7 cosθ₂ - 10/7).
So sinθ₂ = 3.75μ_s(17/7 cosθ₂ - 10/7)
Okay I start to understand a little. In short for sinθ₂ = 3.75μ_s(17/7 cosθ₂ - 10/7) to be defined (which means sliding occurs), θ₂ must be less than 54.0° = θ₁.
Since sinθ₂ > 0,
17/7 cosθ₂ - 10/7 > 0 so cosθ₂ > 10/17 > cosθ₁.
So θ₂ < θ₁.

Okay I think I got it.

 

[PeroK]

Here's an alternative. If $F(\theta)$ is the tangential force and $f(\theta)$ is the frictional force, you can show that, for rolling without slipping:

$F(\theta) = kf(\theta)$ (for some positive constant $k$)

$F$ is strictly increasing, so $f$ must be strictly increasing. Therefore, slipping must occur when $f$ can no longer increase, which must be before the ball loses contact.

It's worth thinking about how powerful mathematics can be when used in this way.

 

[Yoonique]

Why is f_s strictly increasing? I thought f_s = μ_sN. As θ increasing, N actually decreases so f_s decreases.

 

[PeroK]

Why is f_s strictly increasing? I thought f_s = μ_sN. As θ increasing, N actually decreases so f_s decreases.

IF the ball is rolling without slipping, THEN $f$ is increasing. So, IF $f$ is decreasing, THEN the ball is slipping.

 

[Yoonique]

Okay I get it. So it means that the moment the ball rolls down, it is already slipping.

 

[PeroK]

Okay I get it. So it means that the moment the ball rolls down, it is already slipping.

Not quite. $f=\mu N$ is the maximum possible friction force. The friction force will usually be less than this maximum. In fact, for this motion, it will continuously increase from effectively 0.

The ball starts to slip at the point where $f = \mu N$ (as you've already worked out). After that point, $f$ would have to increase to sustain rolling, but it cannot increase any more; and, in fact, it must start to decrease as $N$ continues to decrease.

 

[Yoonique]

Oh I forgot the most crucial condition for static friction which is f_s ≤ μ_sN. Am I right to say that f_s increases from 0 while μ_sN is decreasing as N is decreasing. The moment f_s increases to the value of μ_sN, it starts to slip. Then μ_sN continues to decrease, thus f_s continues to decrease since f_s is at its maximum value of μ_sN.

 

[PeroK]

Oh I forgot the most crucial condition for static friction which is f_s ≤ μ_sN. Am I right to say that f_s increases from 0 while μ_sN is decreasing as N is decreasing. The moment f_s increases to the value of μ_sN, it starts to slip. Then μ_sN continues to decrease, thus f_s continues to decrease since f_s is at its maximum value of μ_sN.

Yes. That's about it.