A sphere lies on a frictionless, horizontal surface. A horizontal force is applied to the sphere (but above or below its center of mass).

  • Will the sphere roll or slide? (I think it will roll based on this answer)
  • If it does roll, then will the acceleration be different compared to if the force is applied to the center of mass?
  • $\begingroup$ This might help - Toppling of a cylinder on a block $\endgroup$
    – mmesser314
    Apr 15 at 15:08
  • $\begingroup$ Since the surface is frictionless, all it does is provide a normal force to counter gravity. So get rid of the surface and put the sphere in outer space to get rid of gravity too. This might help you visualize the situation better, and in particular highlight the fact that that the sphere is free to rotate independently of its linear movement, and that a force applied to the sphere can impart both rotation and linear momentum in varying proportions. $\endgroup$ Apr 15 at 18:09

1 Answer 1


lets write the equations

$$ M\dot v=F\\ I_C\dot\omega=F\,\rho$$

where $~\rho~$ the distance from the center of mass to the force $~F~$

the velocity at the contact point $~v_c~$ is

$$ v_c=\omega\,r-v \tag 1$$


  • $~v_c=0~$ sphere is rolling
  • $~v_c > 0~$ sphere is sliding

from the EOM's

$$ v=\frac{F}{M}\,t\\ \omega=\frac{F\,\rho}{I_C}\,t$$

from here with $~I_C=\frac 25 M\,r^2~$ you obtain

$$ v_c=\frac 52\,\frac \rho r-1,\quad \rho < r$$

thus if $~\rho=\frac 25 r~$ the sphere is rolling and if $~\frac 25 r < \rho < 1~$ the sphere is sliding


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