Minimizing Friction for a Ball in a Rough Bowl: Accelerations and Velocity

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In summary, the ball of mass m and radius r is released from horizontal position and the tangential and radial accelerations and velocities are calculated. The friction coefficient must be minimal at an angle θ so that the ball won't slip.
  • #1
Karol
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Homework Statement


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The ball of mass m and radius r is released from horizontal position, the edge of the bowl.
What are the tangential and radial accelerations at angle θ and what is the velocity.
What should be the minimal friction coefficient at angle θ so that the ball won't slip

Homework Equations


Radial acceleration: ##a_r=\frac{v^2}{r}##

The Attempt at a Solution


Conservation of energy: ##mgh=\frac{1}{2}mv^2~~\rightarrow~~mg(R-r)\sin\theta=\frac{1}{2}mv^2##
$$v^2=2g(R-r)\sin\theta$$
$$a_t=\frac{mg\cos\theta}{m}=g\cos\theta$$
$$a_r=\frac{v^2}{r}=\frac{2g(R-r)\sin\theta}{R}$$
In order not to slip the tangential forces of friction and the component of gravity must cancel:
$$mg\mu=mg\cos\theta~~\rightarrow~~\mu=\cos\theta$$
 
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  • #2
What about rotational KE?
 
  • #3
$$mgh=\frac{1}{2}I_A\omega^2~~\rightarrow~~mg(R-r)\sin\theta=\frac{1}{2}m(k+1)r^2\omega$$
$$\rightarrow~~\omega=\frac{2g(R-r)\sin\theta}{(k+1)r^2},~~v=\omega r$$
$$a_t=\frac{F_t}{m}=\frac{mg\cos\theta}{m}=g\cos\theta$$
$$a_r=\frac{v^2}{R-r}$$
The condition for minimum μ remains the same:
$$mg\mu=mg\cos\theta~~\rightarrow~~\mu=\cos\theta$$
 
  • #4
Karol said:
$$a_t=\frac{F_t}{m}=\frac{mg\cos\theta}{m}=g\cos\theta$$
What about friction?
 
  • #5
At going up and also down the friction acts against gravity's component:
$$F_t-f=mg~~\rightarrow~~mg(\cos\theta-\mu\sin\theta)=ma_t$$
 
  • #6
I was also trying this.

OP: what is R?

And I think the other poster is saying you have to set gravitational potential energy = translational ke and rotational ke. For both radial and translational acceleration. We know translational acceleration will be smaller than if the ball was purely sliding.
 
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  • #7
$$M=I_A\alpha~~\rightarrow~~mrg\cos\theta=mr^2(k+1)\alpha$$
$$\alpha=\frac{g\cos\theta}{(k+1)r},~~v=\omega r$$
R is the bowl's radius.
$$a_t=\alpha r=\left( \frac{\cos\theta}{k+1} \right)g,~~a_r=\frac{v^2}{R}$$
 
  • #8
Karol said:
$$M=I_A\alpha~~\rightarrow~~mrg\cos\theta=mr^2(k+1)\alpha$$
$$\alpha=\frac{g\cos\theta}{(k+1)r},~~v=\omega r$$
R is the bowl's radius.
$$a_t=\alpha r=\left( \frac{\cos\theta}{k+1} \right)g,~~a_r=\frac{v^2}{R}$$

Can't you just set mgh = both types of kinetic energy and solve for v?

This will then give you radial acceleration. v^2/R

Then use kinematics to figure out tangential acceleration
 
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  • #9
$$EK_{(rot)}=\frac{1}{2}I_c\omega^2=\frac{1}{2}I_c\frac{v^2}{r^2}=\frac{1}{2}kmv^2$$
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2=\frac{1}{2}mv^2(k+1)$$
$$mg(R-r)\sin\theta=\frac{1}{2}mv^2(k+1)~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_r=\frac{v^2}{R-r}=\left( \frac{2\sin\theta}{k+1} \right)g$$
How do i use kinematics to find tangential acceleration? Isn't tangential acceleration only a function of the forces (gravity and friction) acting on the ball?
Kinematics deals with velocities, accelerations and distances, not forces. the tangential acceleration acts on the center, right?
The forces depend on the angle, so i can't use constant acceleration formulas.
 
  • #10
Karol said:
$$EK_{(rot)}=\frac{1}{2}I_c\omega^2=\frac{1}{2}I_c\frac{v^2}{r^2}=\frac{1}{2}kmv^2$$
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2=\frac{1}{2}mv^2(k+1)$$
$$mg(R-r)\sin\theta=\frac{1}{2}mv^2(k+1)~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_r=\frac{v^2}{R-r}=\left( \frac{2\sin\theta}{k+1} \right)g$$
How do i use kinematics to find tangential acceleration? Isn't tangential acceleration only a function of the forces (gravity and friction) acting on the ball?
Kinematics deals with velocities, accelerations and distances, not forces. the tangential acceleration acts on the center, right?
The forces depend on the angle, so i can't use constant acceleration formulas.

So to start with your diagram does not have R as a given, only h. Hopefully you were given R as it would make finding centripetal acceleration a lot easier. I also hope the question meant to say a ball of mass m WITH a radius of r? I am having some R, r confusion.

What is your answer for v at the h vertical level? All you need is conservation of energy, no forces or angles required IMO. This v is really just speed in variable form. You can convert omega to v. (omega x r = v) The parenthetical is a key part of your mess up I believe.

And I think if you do know v at h, tangential acceleration is easy at h below the starting vertical level. ( I am assuming the ball starts from rest) You just use the instantaneous speed you found using conservation of energy kinematically. Again, no forces or angles required. Now the last part is going to require forces.
 
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  • #11
pgardn said:
All you need is conservation of energy, no forces or angles required I
The question asks for accelerations, though. I suppose you could use a = v dv/dx, but the direct approach using forces and moments looks easier to me.
 
  • #12
pgardn said:
What is your answer for v at the h vertical level? All you need is conservation of energy, no forces or angles required IMO. This v is really just speed in variable form. You can convert omega to v. (omega x r = v)

The ball starts from the edge of the bowl, with zero speed. What is the normal force and friction at the uppermost position of the ball? Can the ball roll from the beginning? What makes the ball roll ?
If it is not pure rolling, the ball skids at the beginning part of its track, and you can not apply conservation of energy.
 
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  • #13
From post #9:
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_t=\dot{v}=\frac{(R-r)\cos\theta}{(k+1)\sqrt{\frac{2(R-r)}{k+1}\sin\theta}}$$
haruspex said:
but the direct approach using forces and moments looks easier to me.
Do you mean what i wrote in post #7:
$$M=I_A\alpha~~\rightarrow~~\alpha=\frac{g\cos\theta}{(k+1)r},~~\omega=\int \alpha$$
For that i need ##\theta(t)=?##
ehild said:
Can the ball roll from the beginning? What makes the ball roll ?
The normal force at the beginning is zero and the ball cannot roll. the friction makes it roll.
 
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  • #14
Karol said:
The normal force at the beginning is zero and the ball cannot roll. the friction makes it roll.
Then it slides at some part of its track. Can you derive the speed from conservation of energy?
 
  • #15
Karol said:
For that i need θ(t)=?θ(t)=?\theta(t)=?
If work is conserved (see below) you can get the velocities in terms of theta from that, then use the velocities and moments etc. to find acceleration, again as a function of theta. You are not asked to find anything as a function of time, and indeed that might be much harder.
Karol said:
The normal force at the beginning is zero and the ball cannot roll. the friction makes it roll
If there is no normal force, how can there be friction? Yet you need angular acceleration to match the linear acceleration to roll.
 
  • #16
ehild said:
Then it slides at some part of its track. Can you derive the speed from conservation of energy?
But immediately after the beginning there is normal force and it starts rolling, so what i did is correct:
$$mgh=\frac{1}{2}I_A\omega^2~~\rightarrow~~\omega=\frac{2g(R-r)\sin\theta}{(k+1)r^2},~~v=\omega r$$
 
  • #17
Karol said:
But immediately after the beginning there is normal force and it starts rolling, so what i did is correct:
$$mgh=\frac{1}{2}I_A\omega^2~~\rightarrow~~\omega=\frac{2g(R-r)\sin\theta}{(k+1)r^2},~~v=\omega r$$
There is linear acceleration right from the start, so for rolling contact there should be rotational acceleration from the start. But since there is no normal force then there is no frictional torque to provide it. In order to make the later transition to rolling contact, it will then have to go through a sliding phase, in which kinetic friction brings it up to rotational speed.
 
  • #18
ehild said:
Then it slides at some part of its track. Can you derive the speed from conservation of energy?
Sliding makes things worse since i have to add friction losses which i don't know:
$$mgh=\frac{1}{2}I_A\omega^2+E_f$$
 
  • #19
ehild said:
The ball starts from the edge of the bowl, with zero speed. What is the normal force and friction at the uppermost position of the ball? Can the ball roll from the beginning? What makes the ball roll ?
If it is not pure rolling, the ball skids at the beginning part of its track, and you can not apply conservation of energy.
Holy cow I never took that into account. The last part of the question now makes a lot more sense. It's a rolling and slipping problem. I should have known from the last question.

Thanks.

Very sorry Karol.

I will just observe.
 
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  • #20
haruspex said:
There is linear acceleration right from the start, so for rolling contact there should be rotational acceleration from the start. But since there is no normal force then there is no frictional torque to provide it. In order to make the later transition to rolling contact, it will then have to go through a sliding phase, in which kinetic friction brings it up to rotational speed.

Thanks for this addition also.

Sorry Karol.
 
  • #21
To find where pure rolling starts:
$$M=I_A\alpha~~\rightarrow~~\alpha=\frac{g\cos\theta}{(k+1)r},~~\omega=\int_0^{\theta} \alpha=\left( -\frac{g\mu}{kr^2} \right)\cos\theta$$
$$F_t-f=ma_t:~~mg\cos\theta-mg\mu\sin\theta=ma_t$$
$$\rightarrow~~v=\int\frac{F_t}{m}=g\int_0^{\theta}(\cos\theta-\mu\sin\theta)=-g(\sin\theta+\mu\cos\theta)$$
The dimensions aren't correct, and what does the minus sign mean in ω and v?
 
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  • #22
ehild said:
Then it slides at some part of its track. Can you derive the speed from conservation of energy?
@ehild, can you see a way to solve the first part? Eliminating time, I get an equation of the form ##yy'+ky^2=A\cos(x)-B\sin(x)##. The equation with time has ##\theta, \dot\theta, \ddot\theta## in it, so there seems to be no way to find the accelerations without solving it.
 
  • #23
Karol said:
To find where pure rolling starts:
$$M=I_A\alpha~~\rightarrow~~\alpha=\frac{g\cos\theta}{(k+1)r},~~\omega=\int_0^{\theta} \alpha=\left( -\frac{g\mu}{kr^2} \right)\cos\theta$$
$$F_t-f=ma_t:~~mg\cos\theta-mg\mu\sin\theta=ma_t$$
$$\rightarrow~~v=\int\frac{F_t}{m}=g\int_0^{\theta}(\cos\theta-\mu\sin\theta)=-g(\sin\theta+\mu\cos\theta)$$
The dimensions aren't correct, and what does the minus sign mean in ω and v?
I don't understand the bit after ##\int_0^{\theta}\alpha##. How are you integrating that with respect to time without knowing alpha as a function of time?
 
  • #24
haruspex said:
How are you integrating that with respect to time without knowing alpha as a function of time?
I don't know if i integrate with respect to time or θ. i hoped that if α=α(θ) then i need to integrate with respect to θ.
@haruspex, how did you get ##yy'+ky^2=A\cos(x)-B\sin(x)## and what is x, is it the angle θ?
 
  • #25
haruspex said:
@ehild, can you see a way to solve the first part? Eliminating time, I get an equation of the form ##yy'+ky^2=A\cos(x)-B\sin(x)##. The equation with time has ##\theta, \dot\theta, \ddot\theta## in it, so there seems to be no way to find the accelerations without solving it.
During the sliding part of the motion, the accelerations a and α can be expressed with the forces:
ma=mgcos(θ) - μN
kmr2α=rμN.
Knowing N(θ), we know the accelerations.
Solving the equation ##yy'+ky^2=A\cos(x)-B\sin(x)## for y=(dθ/dt) in terms of x=theta, we find N=mv2 /(R-r) + mg sin(θ).
 
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  • #26
Karol said:
i hoped that if α=α(θ) then i need to integrate with respect to θ.
##\alpha=\dot\omega##, the derivative wrt time, so to get ω from α it would have to be an integral wrt time.
Karol said:
what is x, is it the angle θ
Yes, and y is ω. y' is dy/dx. But the k I wrote is not your k, just some constant.
 
  • #27
ehild said:
y=(dθ/dt)2
I don't think that's quite what you meant, but yes, I see how to solve it now, thanks. (Doh!)
ehild said:
we find N= ...
No, that's where I came from to get my differential equation. To find N we need to find v, so need to find omega...
The solution I get now is ##\dot\theta^2=\frac g{(R-r)(1+4\mu_k^2)}[6\mu_k(\cos(\theta)-e^{-2\mu_k\theta})+\sin(\theta)]##
Wow. Can it really be that complicated? Can't find a mistake in my working.
 
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  • #28
haruspex said:
The solution I get now is ##\dot\theta^2=\frac g{(R-r)(1+4\mu_k^2)}[6\mu_k(\cos(\theta)-e^{-2\mu_k\theta})+\sin(\theta)]##
I get a very similar solution except for the coefficient of sinθ. Where you have sinθ, I get (1-2μ2)sinθ. But I could be mistaken.
 
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  • #29
TSny said:
I get a very similar solution except for the coefficient of sinθ. Where you have sinθ, I get (1-2μ2)sinθ. But I could be mistaken.
You are right, I dropped a factor.
 
  • #30
I wonder if the problem wants you to assume (unrealistically) that the ball rolls without slipping from the beginning. Then, for an arbitrary value of θ, find the minimum coefficient of static friction that will prevent slipping at that θ. You can then see how ##\mu^{\rm min}_s## diverges as θ → 0.

Probably not going to get many to concur with this view.
 
  • #31
TSny said:
I wonder if the problem wants you to assume (unrealistically) that the ball rolls without slipping from the beginning. Then, for an arbitrary value of θ, find the minimum coefficient of static friction that will prevent slipping at that θ. You can then see how ##\mu^{\rm min}_s## diverges as θ → 0.

Probably not going to get many to concur with this view.
The problem with that is that to figure out if it will commence rolling at some theta we need to calculate the acquired rotation to that point, i.e. while sliding. So we are back having to use that rebarbative equation.
Assuming it doesn't slip until then doesn't work. If it hasn't slipped earlier, it is not going to start slipping at theta.
 
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  • #32
haruspex said:
The problem with that is that to figure out if it will commence rolling at some theta we need to calculate the acquired rotation to that point, i.e. while sliding. So we are back having to use that rebarbative equation.
Assuming it doesn't slip until then doesn't work. If it hasn't slipped earlier, it is not going to start slipping at theta.
Yes, I see your point.
 
  • #33
haruspex said:
I don't think that's quite what you meant, but yes, I see how to solve it now, thanks. (Doh!)
Yes, I corrected it. I was a bit ahead, as we can make a first order equation for (dθ/dt)2.
haruspex said:
No, that's where I came from to get my differential equation. To find N we need to find v, so need to find omega...
You mean to find (dθ/dt). But you found it if you solved the de for y.
 
  • #34
TSny said:
I wonder if the problem wants you to assume (unrealistically) that the ball rolls without slipping from the beginning. Then, for an arbitrary value of θ, find the minimum coefficient of static friction that will prevent slipping at that θ. You can then see how ##\mu^{\rm min}_s## diverges as θ → 0.

Probably not going to get many to concur with this view.
I thought the same :) No problem makers are correct. And the text of the problem is not quite clear.
 
  • #35
ehild said:
You mean to find (dθ/dt). But you found it if you solved the de for y.
OK, I was confused by your writing "solving the equation ... we find ...". You meant, solving the equation, we can plug the velocity into ... and hence find N.
 

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