Rotational motion inclined plane

[pff]

Hi guys,
I need to model a ball rolling on an incline plane, and i would like to be able to calculate acceleration due to gravity for a given angle.
Currently i have:
a = f / m
f = m*g*sin(angle)
a = g*sin(angle)

I am aware this does not take rotational motion into account, which is what I'm trying to do now
Its been a while since i took a physics class, If someone could help me out it would be much appreciated

 

[Doc Al]

When you have the ball rolling without slipping on an incline, you'll need to include the force of friction. By applying Newton's laws to translation and rotation, and the constraint for rolling without slipping (v = ωr), you'll be able to figure out the acceleration as a function of angle.

 

[pff]

ok, i came up with the following

a = $$\alpha$$r
$$\alpha$$ = $$\tau$$/I
solid sphere I = (2/5)mr$$^{2}$$
$$\tau$$=rF
$$\alpha$$ = rF/(2/5)mr$$^{2}$$
a = Fr/(2/5)mr

= g*sin(angle) / (2/5)

Is this sensible? i would have expected an r term.

 

[Doc Al]

Almost, but not quite.

You need to apply Newton's 2nd law to both rotation and translation. You'll end up with two equations, which you'll combine.

 

[pff]

Translational motion due to the ball slipping?
Now my head is starting to hurt.
I would need to have a value for friction, then split the force between rotational acceleration and acceleration down the plane?
I don't have any specific value for friction coefficient.
If there is sufficient friction could I just ignore the translational component?

 

[Doc Al]

Translational motion due to the ball slipping?

No, translational motion due to the ball rolling down the incline. (It translates as well as rotates.

Now my head is starting to hurt.
I would need to have a value for friction, then split the force between rotational acceleration and acceleration down the plane?
I don't have any specific value for friction coefficient.
If there is sufficient friction could I just ignore the translational component?

You don't need to know the value of the friction, just call it F. (No, you can't ignore translational motion!)

What forces act on the ball? Draw a free body diagram.

 

[pff]

well i already have translational acceleration
a = g*sin(angle)
...can i just add them together?
a = a(rot) + a(tran) = g*sin(angle) / (2/5) + g*sin(angle)

 

[Doc Al]

well i already have translational acceleration
a = g*sin(angle)

That's the translational acceleration if the ball slid down the incline without rolling. (What would happen with no friction, for example.) When the ball rolls, things change.

...can i just add them together?
a = a(rot) + a(tran) = g*sin(angle) / (2/5) + g*sin(angle)

No. Do what I said earlier. Apply Newton's 2nd law twice.

 

[pff]

I guess i have to account for friction then?

a = m*g*sin(angle) - F / m

Your saying apply Newtons laws twice? i suppose this is to get a net force?

F(tran) = m*g*sin(angle) - F(friction)

F(rot) = ma = m*g*sin(angle)/(2/5)

F = m*g*sin(angle) - F(friction) + m*g*sin(angle)/(2/5)

Is that what you had in mind? along the right track?
I doubt it because using F=ma the m cancels and its equivalent to summing acceleration. I hope you just made me do properly, because I'm stuck otherwise, I need more of a hint please.

 

[Doc Al]

I guess i have to account for friction then?

Of course.

a = m*g*sin(angle) - F / m

Almost, but not quite.

Your saying apply Newtons laws twice? i suppose this is to get a net force?

No, not to get a net force. Once for rotation (where net torque is used instead of net force) and once for translation.

F(tran) = m*g*sin(angle) - F(friction)

OK, that's the net force. Set it equal to ma:
m*g*sin(angle) - F(friction) = ma

F(rot) = ma = m*g*sin(angle)/(2/5)

F = m*g*sin(angle) - F(friction) + m*g*sin(angle)/(2/5)

Is that what you had in mind? along the right track?
I doubt it because using F=ma the m cancels and its equivalent to summing acceleration. I hope you just made me do properly, because I'm stuck otherwise, I need more of a hint please.

Hint: Newton's 2nd law for rotation is Torque = I*alpha. (You used it earlier.)

What torque does friction exert on the ball? How does alpha relate to a?

 

[pff]

Sorry, a = g*sin(angle) - F / m

I thought torque from friction was equivalent to the force pulling the ball down the slope in order to make it rotate

torque = m*g*sin()*r
alpha = a/r, but i don't understand where that gets me, back to a = g*sin(angle) / (2/5) ?

 

[Doc Al]

Sorry, a = g*sin(angle) - F / m

OK, but you're getting ahead of yourself. So far, we have one equation, for translation. (See last post.)

I thought torque from friction was equivalent to the force pulling the ball down the slope in order to make it rotate

torque = m*g*sin()*r
alpha = a/r, but i don't understand where that gets me, back to a = g*sin(angle) / (2/5) ?

Not sure what you're doing here. Torque due to friction is just F*r. So write Newton's 2nd law for rotation. That will give you your second equation.

 

[pff]

sorry, so i have

F = m*g*sin(angle) - F(friction)
and
t = rF = r*m*g*sin(angle)

...?

 

[Doc Al]

sorry, so i have

F = m*g*sin(angle) - F(friction)

I was using "F" to represent friction; Instead, for clarity, let's call Friction F_f.
The net force is:
ΣF = m*g*sin(angle) - F_f
Apply Newton's 2nd law:
m*g*sin(angle) - F_f = ma
[this is equation #1]

and
t = rF = r*m*g*sin(angle)

The torque producing force is the friction; it's torque is r*F_f. (Nothing to do with gravity.)

Apply Newton's 2nd law:
Torque = I*alpha

Flesh that last equation out; it will be equation #2.

 

[pff]

ok, but if i have

t = I$$\alpha$$

so
I = (2/5)*m*r^2
but angular acceleration is due to the angular component of gravity, although the friction causes the rolling, the amount is due to g isn't it?
...
unless i use a = $$\alpha$$r
$$\tau$$ = I*a/r
is that better?

 

[Doc Al]

ok, but if i have

t = I$$\alpha$$

so
I = (2/5)*m*r^2
but angular acceleration is due to the angular component of gravity, although the friction causes the rolling, the amount is due to g isn't it?

You're thinking way too much. The torque is due to the friction, which depends on the angle. But let the equations do the work for you.

...
unless i use a = $$\alpha$$r
$$\tau$$ = I*a/r
is that better?

Yes. So far, so good. Replace tau and I with what they equal.

 

[pff]

t = I*a/r
= (2/5)*m*r^2*a / r

t = Ff*r
Ff*r = (2/5)*m*a*r

F = (2/5)*m*a
the force? that's what we want first right?

 

[Doc Al]

t = I*a/r
= (2/5)*m*r^2*a / r

t = Ff*r
Ff*r = (2/5)*m*a*r

F = (2/5)*m*a
the force? that's what we want first right?

Yes! (Make sure that F is really F_f, the friction force.)

This is equation #2. Combine it with equation #1, and then solve for the acceleration. (You can also solve for the friction force, if you want.)

 

[pff]

Ff = (2/5)*m*a
ma = m*g*sin(t) - Ff

ma = m*g*sin(t) - (2/5)*m*a
a = g*sin(t) - (2/5)*a
a + (2/5)*a = g*sin(t)
(7/5)*a = g*sin(t)

a = g*sin(t) / (7/5)

..?
please?

 

[Doc Al]

Ff = (2/5)*m*a
ma = m*g*sin(t) - Ff

ma = m*g*sin(t) - (2/5)*m*a
a = g*sin(t) - (2/5)*a
a + (2/5)*a = g*sin(t)
(7/5)*a = g*sin(t)

a = g*sin(t) / (7/5)

..?
please?

Yes! You made it!

But write it as
a = (5/7)*g*sin(t)

 

[pff]

woohoo!

Thanks for your time and patience, its very much appreciated.

 

[Doc Al]

You're welcome.

Notice that the acceleration is less than something sliding with no friction. This should make sense, since some of the gravitational PE is being converted to rotational KE, leaving less translational KE.

But no energy is lost to friction, since there's no slipping. Mechanical energy will be conserved.