Angular Acceleration & Friction Force of Rolling Cylinder

[Jayhawk1]

No one in my class seems to get this:

A 4.8-kg uniform cylinder rolls without slipping down a plane at 26.3-degree angle above horizontal. (a) What is the translational acceleration of the cylinder? (b) What is the magnitude of the frictional force?

Now I know that the without slipping part is a crucial part of the problem that dictates what forumula should be used. I don't know what formula to use exactly though, because none of them seem to fit.

 

[whozum]

Translational motion is talking about motion external to the actual object. In this case, translational motion means the motion of the cylinder neglecting any spinning or rotating effects.

Drawing a force diagram, identify the forces acting on the cylinder, and find the magnitude of this force at the angle given. Divide the force by mass.

For part b, recall the frictional force is $$F_{friction} = \mu N$$ where N is the normal force $$mgcos(\theta)$$. Other than that I've got nothing for part b.

 

[Jayhawk1]

Ok, I understand

 

[marlon]

Translational motion is talking about motion external to the actual object. In this case, translational motion means the motion of the cylinder neglecting any spinning or rotating effects.

I really think the original poster meant to say tangential acceleration.

Suppose the object with rotational inertia I, rolls down from an incline. We need to apply the two versions of Newton's second law. We have gravity, friction and a normal force. Also, for the torque, it is important to realize that the friction-force-vector is perpendicular to the radius at the point where the object touches the ground. Suppose that the x-axis is directed upwards, so the object rolls in opposite x-direction.

The object moves opposite the x-direction, so a (tangential acceleration along the x-axis) will be negative

The direction of our rotation is clockwise so the torque and the angular acceleration (alpha) should be negative.

$$F_{friction} -Mgsin( \theta ) = Ma$$
$$\tau = -F_{friction}R = I \alpha$$

and

$$alpha = \frac{a}{R}$$ where a is the tangential acceleration

Now just solve the above equations for a and your problem is okelidokeli...

You should get $$a = \frac{-gsin( \theta )}{1 + \frac{I}{MR^2}}$$

R is the radius of the object, M is the mass

regards
marlon

 

[Jayhawk1]

...But I wasn't given a radius...

 

[marlon]

...But I wasn't given a radius...

Hmm, ok then, i gave too much info...Forget about the rotation of the object. Just keep on working with the first version of Newton's second law...sorry

regards
marlon

 

[Jayhawk1]

Ok, I'll try that. Thank you.

 

[Jayhawk1]

I'm so sorry... but I still am not getting the right answer. I don't know what I am doing wrong.

 

[whozum]

For part a)

$$Mass = 4.8kg$$

$$Acceleration = 9.8 ms^-2*sin(26.3) = 4.34 ms^-2$$

$$Force = Mass x Acceleration$$

That should be the answer to a. If you had more information I could help you with b, but I can't think of any way to find the frictional force without a coefficient of friction.

 

[Jayhawk1]

That is not the correct answer I tried that the very first time and the computer told me I was wrong. I don't understand, it should be simple, considering it is only the translational accl.

 

[marlon]

For part a)

$$Mass = 4.8kg$$

$$Acceleration = 9.8 ms^-2*sin(26.3) = 4.34 ms^-2$$

$$Force = Mass x Acceleration$$

That should be the answer to a. If you had more information I could help you with b, but I can't think of any way to find the frictional force without a coefficient of friction.

x-axis is pointing up the incline

you are forgetting the normal force.

$$\mu N - mgsin( \theta) = ma$$

$$N -mgcos( \theta ) = 0$$

solve the last equation for N and plug it in the first one that you solve for a

marlon

 

[whozum]

Perhaps I'm not understanding the original problem. Is it not a standard rolling down ramp problem? Is the not slipping acting as a restriction to the acceleration?

 

[Jayhawk1]

I think my Professor isn't the best at creating these questions... because I don't have a coeff. of friction, right?

 

[Jayhawk1]

Marlon, the funny thing is that these questions were made by the Physics department chair at the University of Kansas, and my whole class doesn't seem to understand.

 

[whozum]

Yes, you have two unknowns, your resultant acceleration and your friction coefficient.

 

[Yegor]

You definitely should not forget about rolling.
Use
$$a = \frac{-gsin( \theta )}{1 + \frac{I}{MR^2}}$$

Do you know I for uniform cylinder?

 

[marlon]

You definitely should not forget about rolling.
Use
$$a = \frac{-gsin( \theta )}{1 + \frac{I}{MR^2}}$$

Do you know I for uniform cylinder?

Yes, that was my first hint too but indeed you CAN treat the object not as a solid object but as a 'point-particle' , as is often done...

marlon

 

[Yegor]

Marlon's 4th post was absolutely right (I suppose).
For uniform cylinder I=(M*R^2)/2
Thus: a=(2g/3)*Sin(theta)

 

[marlon]

Marlon's 4th post was absolutely right (I suppose).
For uniform cylinder I=(M*R^2)/2
Thus: a=(2g/3)*Sin(theta)

That is indeed how i would solve it too

marlon

 

[Integral]

The only force acting to cause rotation is the frictional force so we have a torque $\tau$ applied.

From the kenimatics of rotation we have

$$\tau = I_{cm} \alpha$$

The torque due to friction is

$$\tau = f R$$

For a cylinder

$$I_{cm} = \frac 1 2 MR^2$$

combining these and using $\alpha = \frac a R$ we get

$$f = \frac 1 2 Ma$$

Now applying Newtons Law to the motion down the incline we have
$$Mg \sin( \theta ) - f = M a$$

we can now solve for a

$$a = \frac 2 3 g sin ( \theta )$$

Of course $\theta$ is the angle of your incline.

 

[plusaf]

and notice...

The only force acting to cause rotation is the frictional force so we have a torque $\tau$ applied.

From the kenimatics of rotation we have

$$\tau = I_{cm} \alpha$$

The torque due to friction is

$$\tau = f R$$

For a cylinder

$$I_{cm} = \frac 1 2 MR^2$$

combining these and using $\alpha = \frac a R$ we get

$$f = \frac 1 2 Ma$$

Now applying Newtons Law to the motion down the incline we have
$$Mg \sin( \theta ) - f = M a$$

we can now solve for a

$$a = \frac 2 3 g sin ( \theta )$$

Of course $\theta$ is the angle of your incline.

notice that the actual coefficient of static friction was not needed... if the cylinder is rolling without slipping, the coefficient was high enough to prevent slipping. beyond that, it can be any higher number...