3
$\begingroup$

I came across a question which says:

"A force $F$ acts tangentially at the highest point of a sphere of mass m kept on a rough horizontal plane. If the rolls without slipping , find the acceleration of the centre of mass of the sphere."

The equations in the solution are:

For translational motion,

$F+f=ma$

For rotational motion about centre,

$Fr-fr=I{\alpha}$

where $f$ represents frictional force, $a$ is linear acceleration of the sphere and $\alpha$ is angular acceleration of the sphere, $r$ is the radius of the sphere.

At first I thought that the surface would provide friction to the left side for bothe translation and rotation of the sphere, but was only amazed to look at the reason given in the solution.

The reason is given as -

"As the force $F$ rotates the sphere, the point of contact has a tendency to slip towards left so that the static friction on the sphere will act towards right"

Is it the correct explanation?

Also I wanted to make sure whether tangential force always tends to slip a sphere in opposite direction and static friction acts towards direction of that tangential force.

$\endgroup$
2
  • $\begingroup$ you can take it as a formula, if height of point of application of external force is above $h=I/mr$, where I is moment of inertia, friction force is parallel to external force, $\endgroup$
    – PinkAura
    Commented Mar 18 at 7:16
  • $\begingroup$ you can very easily derive it yourself (hint: find height $h$ for which there is no friction but rolling without slipping occurs) $\endgroup$
    – PinkAura
    Commented Mar 18 at 7:18

2 Answers 2

2
$\begingroup$

YES, that explanation is correct.

The direction of friction during rolling down a wedge is a very common misconception.

However, it is very simple if we concentrate of the concept of friction.

FRICTION is an electromagnetic force between two particle IN DIRECT CONTACT. Its only job is to PREVENT SLIPPING.

Since it

  1. Prevents slipping
  2. Acts only on the particle directly in contact with frictional surface

Thus, friction acts against the direction of acceleration (not necessarily in direction of velocity) of the particle in contact with the wedge

Now let me explain it using an example, Assume we have a wheel and a stick attached to its center. Now pull the stick horizontally with a force F F = ma Thus, direction of acceleration is same as direction of force. Now, the particle in contact with the ground most be at rest since the ground is at rest, so, to give the point of contact an equal and opposite acceleration friction acts. Now, assume a similar case but this time another force 'f' is applied to the topmost point, this f cause torque = fr and hence providing angular acceleration. IF this angular acceleration is greater than a/r (initial angular acceleration) then, the point of contact starts slipping backwards and hence friction acts FORWARD this time.

I hope this helps

$\endgroup$
2
$\begingroup$

I don't think that your explanation is correct

enter image description here

look at the FBD . you can move the force F to the center of mass ,thus you obtain torque $~F\,r~$. The force F cause the sphere to move to the right , so I put the velocity v to the right and the friction force $~F_\mu~$ opposite to the direction of the velocity.

the equations are:

$$m\,a=F-F_\mu\\ I\,\alpha=F\,r+F_\mu\,r$$

now those equations don't work, if F change sign. make it work for any sign of the external force F, the friction force

$$F_\mu\mapsto F_\mu\,\rm sign(v)=F_\mu\frac{v}{|v|} $$

Numerical simulation

enter image description here

enter image description here

$\endgroup$
6
  • $\begingroup$ Why are you justified in moving the application point of F? In the original problem, F may provide a torque that causes the sphere to roll, but in this formulation, friction is the only thing that acts off-axis and can apply a torque. In the original problem, the sphere would rotate whether there was friction or not; here it would not rotate if there were no friction. I'm not sure that reasoning about what happens when applying F to the CoM tells us much about applying F elsewhere. $\endgroup$ Commented Mar 18 at 13:06
  • $\begingroup$ The froce F is Not tied together with the sphere? $\endgroup$
    – Eli
    Commented Mar 18 at 16:21
  • $\begingroup$ F is defined as acting at the top edge of the sphere. You seem to be arguing it's equivalent to apply F to the center of mass. $\endgroup$ Commented Mar 18 at 16:44
  • $\begingroup$ "You seem to be arguing it's equivalent to apply F to the center of mass" plus torque at the center of mass $~\tau=F\,r~$ you can move the external force where ever you want to, you get the force plus torque. so what is wrong with this ? $\endgroup$
    – Eli
    Commented Mar 18 at 17:01
  • $\begingroup$ The ball's behavior completely changes if you move where F is applied. In a scenario with no friction, the ball will roll forward if you push the top edge, but it will just glide without rolling if you push the CoM. If you push the bottom edge, it will move forward with backspin. Certainly those can't be equivalent. $\endgroup$ Commented Mar 18 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.