Rolling Ball and angle of incline

[frazdaz]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

Moments about the tipping point:
$$mgrsin\theta = \frac{2mr^2}{5}\alpha$$
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

 

[Doc Al]

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

 

[frazdaz]

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

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

 

[Doc Al]

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

Yes, exactly.

 

[frazdaz]

Yes, exactly.

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?
$$\theta = sin^{-1}(\frac{7r}{25})$$

 

[Tanya Sharma]

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?
$$\theta = sin^{-1}(\frac{7r}{25})$$

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

 

[frazdaz]

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

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?

 

[Tanya Sharma]

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?

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

 

[Doc Al]

I've just never seen that before.

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

 

[Doc Al]

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?

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

$$a = \alpha r$$

 

[frazdaz]

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

$$a = \alpha r$$

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

 

[Tanya Sharma]

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

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

 

[Doc Al]

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

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

 

[frazdaz]

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

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.

 

[Doc Al]

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.

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 $a = \alpha r$.