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Suppose a wheel of mass $M$ and radius $R$ is at rest. (Both wheel and surface are assumed to be rigid.) Now I am applying a force $F$ at the center of mass.

  1. What is the minimum force required to start moving the wheel?
  2. Once the wheel starts to roll, what is the friction at the point of contact? Is it static or dynamic?
  3. How can I calculate the velocity after time $t$?
  4. Instead we instead of force apply torque, what will the difference be in the above?

Now I assume both wheel and surface to be real (not rigid).

  1. What is the minimum force required to start moving the wheel in this case?
  2. After a time $t$ I remove my force $F$. How long will it take for it to come to rest?

I will here summarize my efforts to solve these doubts:

  1. Minimum force required to start moving the object will be from a static frictional force.
  2. Once the object starts rolling (assuming without slipping), friction will be dynamic and will act in the opposite direction of motion at the point of contact. The net force is $$F_{net} = F-f_d$$
  3. To calculate my velocity after time $t$, I do: $$V_{cm}=a_{cm}*t \quad \text{where}\quad a_{cm}=F_{net}/M$$

When applying torque, the minimum torque required is $T = f_s*R$. When applying torque, static friction will cause the object to move forward with a velocity and also it will cause angular deceleration. The net torque is then $$T_{net} = T_{applied} -f_s*R$$

Angular acceleration is $\alpha = T_{net}/I$. From this, $a_{cm}$ and $v_{cm}$ can be calculated.

My doubt here is that in both cases, velocity will keep on increasing. At what velocity will slip between wheels and surface occur? Or will no slip occur?

  1. The minimum force required will be $F_r=C*N$, where C is the rolling resistance coefficient, which depends on materials (wheel & surface) and geometry of the wheel.
  2. When force $F$ is removed let us say the velocity is v. Then, only rolling resistance acts, and: $$0=v-a*t\quad \text{where}\quad a=F_r/M$$

Is my understanding correct? Please correct me where appropriate.

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closed as off-topic by Kyle Kanos, Yashas, John Rennie, Jon Custer, peterh May 16 '17 at 2:51

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  1. The friction force is not causing any translational motion here. The force applied to the centre is enough to make the whole wheel move - friction or not. And since no other forces counteract this force, the smallest push of this force will make it move. It just must be more than zero, $F>0$.

  2. You write "assuming without slipping", but then "friction will be dynamic". If you have rolling without slipping, then the contact point is stationary. There is no sliding over the surface, so no kinetic/dynamic friction. Friction on stationary surfaces is called static, and it makes sure that these surfaces that try to slide don't.

  3. You started out good, but unfortunately the friction is not dynamic as you assumed, but static, $F_{net}=F-f_{s}$. You can still use your equations, though, you just need one more equation to find $f_s$. And that extra equation might be $\tau_{net}=I\alpha$ as well as $\alpha R=a_{cm}$.

  4. Correct description. You could have used this same description to solve questions 1. to 3.; the static friction just points in another direction.

My doubt here is " In both cases, velocity will keep on increasing. At what velocity, slip occurs between wheels and surface or no slip occurs?"

Slip will not occour just by increasing the speed. Only if the acceleration is increases because the force (or torque) applied increases, will the static friction have to increase as well to cause the expected angular acceleration. And since there is a maximum limit to static friction $f_s<\mu_s n$, slip will happen when the necessary static friction is larger than this value (when the surfaces are not able to cause the necessary $f_s$).

  1. True, the rolling resistance is the combination of all energy losses (damping effects) from deformations, heat losses, etc. If you know that value, then yes, the real-life rolling friction (or rolling resistance) can be calculated - but that value is only rarely known.

  2. Correct.

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  • $\begingroup$ @SurendarKumar Thank you or the updated question. I have now updated this answer to fit your question. $\endgroup$ – Steeven May 15 '17 at 14:26

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