# Rolling Ball and angle of incline

1. Dec 22, 2013

### frazdaz

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

A uniform solid sphere rolls down an incline without slipping. If the
linear acceleration of the centre of mass of the sphere is 0.2g, then what
is the angle the incline makes with the horizontal? Repeat for a thin
spherical shell.

2. Relevant equations
$$\sum \tau = I \alpha$$
$$I_{ball} = \frac{2mr^2}{5}$$

3. The attempt at a solution
$$mgrsin\theta = \frac{2mr^2}{5}\alpha$$
Don't think this is going anywhere but I can't think of any other way to include moments of inertia.

Just thought, this moment of inertia is from the centre of the ball, not the tipping point. Correct?

So, by the parallel axis theorem,
$$mgrsin\theta = (\frac{2mr^2}{5} + mr^2)\alpha = \frac{7mr^2}{5}\alpha$$

Which hardly helps

Last edited: Dec 22, 2013
2. Dec 22, 2013

### Staff: Mentor

So far, so good. Now relate the linear acceleration of the center of mass to the angular acceleration.

3. Dec 23, 2013

### frazdaz

Like this? (found from one of your posts)
$$a_t= \alpha r$$

Last edited: Dec 23, 2013
4. Dec 23, 2013

### Staff: Mentor

Yes, exactly.

5. Dec 23, 2013

### frazdaz

I've just never seen that before.

$$mgrsin\theta = \frac{7mr^2}{5} \frac{g}{5}$$
$$sin\theta = \frac{7r}{25}$$
Does it want it in terms of r or have I gone wrong?
$$\theta = sin^{-1}(\frac{7r}{25})$$

6. Dec 23, 2013

### Tanya Sharma

Recheck your calculations.You have missed an 'r' while making substitution. 'r' will cancel out .

7. Dec 23, 2013

### frazdaz

Ahhh, of course. Thank you!
I take it the tangential acceleration is in the direction of the linear velocity? And is the radial acceleration just the centripetal acceleration?

8. Dec 23, 2013

### Tanya Sharma

Tangential acceleration and radial acceleration of what ? In this problem we are dealing with linear and angular acceleration of the sphere.

9. Dec 23, 2013

### Staff: Mentor

That's the condition for 'rolling without slipping'.

10. Dec 23, 2013

### Staff: Mentor

The condition for rolling without slipping relates the linear acceleration of the center of mass to the angular acceleration:

$$a = \alpha r$$

11. Dec 23, 2013

### frazdaz

Does that work as an inequality? i.e. object won't slip providing as long as this holds true $$a \leq \alpha r$$

12. Dec 23, 2013

### Tanya Sharma

No...If $a \neq \alpha r$ ,then slipping occurs .For rolling without slipping $a = \alpha r$

13. Dec 23, 2013

### Staff: Mentor

No, it's an equality. If $a \ne \alpha r$, then there is slipping.

14. Dec 23, 2013

### frazdaz

How would it slip if the linear acceleration smaller than the RHS? I can picture the ball accelerating so quickly that it begins to slip, but not the other way around.

15. Dec 23, 2013

### Staff: Mentor

Imagine the ball spinning, without enough traction.

In practice, if you were to start the ball from rest at the top of the incline and there were insufficient friction to prevent slipping, you would have $a \gt \alpha r$.

Nonetheless, if you have rolling without slipping, then $a = \alpha r$.