# Ball rotates in a rough bowl

1. Aug 14, 2016

### Karol

1. The problem statement, all variables and given/known data

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

2. Relevant equations
Radial acceleration: $a_r=\frac{v^2}{r}$

3. 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:
$$mg\mu=mg\cos\theta~~\rightarrow~~\mu=\cos\theta$$

2. Aug 14, 2016

### haruspex

3. Aug 15, 2016

### Karol

$$mgh=\frac{1}{2}I_A\omega^2~~\rightarrow~~mg(R-r)\sin\theta=\frac{1}{2}m(k+1)r^2\omega$$
$$\rightarrow~~\omega=\frac{2g(R-r)\sin\theta}{(k+1)r^2},~~v=\omega r$$
$$a_t=\frac{F_t}{m}=\frac{mg\cos\theta}{m}=g\cos\theta$$
$$a_r=\frac{v^2}{R-r}$$
The condition for minimum μ remains the same:
$$mg\mu=mg\cos\theta~~\rightarrow~~\mu=\cos\theta$$

4. Aug 15, 2016

### haruspex

5. Aug 15, 2016

### Karol

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. Aug 15, 2016

### pgardn

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.

Last edited: Aug 15, 2016
7. Aug 15, 2016

### Karol

$$M=I_A\alpha~~\rightarrow~~mrg\cos\theta=mr^2(k+1)\alpha$$
$$\alpha=\frac{g\cos\theta}{(k+1)r},~~v=\omega r$$
$$a_t=\alpha r=\left( \frac{\cos\theta}{k+1} \right)g,~~a_r=\frac{v^2}{R}$$

8. Aug 15, 2016

### pgardn

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

Last edited: Aug 15, 2016
9. Aug 15, 2016

### Karol

$$EK_{(rot)}=\frac{1}{2}I_c\omega^2=\frac{1}{2}I_c\frac{v^2}{r^2}=\frac{1}{2}kmv^2$$
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2=\frac{1}{2}mv^2(k+1)$$
$$mg(R-r)\sin\theta=\frac{1}{2}mv^2(k+1)~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_r=\frac{v^2}{R-r}=\left( \frac{2\sin\theta}{k+1} \right)g$$
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. Aug 15, 2016

### pgardn

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 gonna require forces.

Last edited: Aug 15, 2016
11. Aug 16, 2016

### haruspex

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. Aug 16, 2016

### ehild

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.

13. Aug 16, 2016

### Karol

From post #9:
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_t=\dot{v}=\frac{(R-r)\cos\theta}{(k+1)\sqrt{\frac{2(R-r)}{k+1}\sin\theta}}$$
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 $\theta(t)=?$
The normal force at the beginning is zero and the ball cannot roll. the friction makes it roll.

Last edited: Aug 16, 2016
14. Aug 16, 2016

### ehild

Then it slides at some part of its track. Can you derive the speed from conservation of energy?

15. Aug 16, 2016

### haruspex

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.
If there is no normal force, how can there be friction? Yet you need angular acceleration to match the linear acceleration to roll.

16. Aug 16, 2016

### Karol

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. Aug 16, 2016

### haruspex

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. Aug 16, 2016

### Karol

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. Aug 16, 2016

### pgardn

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.

Last edited: Aug 16, 2016
20. Aug 16, 2016