A force `F` acts tangentially at the highest point of a sphere `f` mass m kept on a rough horizontal plane. If the sphere rolls without slipping, find the acceleration of the centre of the sphere.
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The situation is shown fig. As 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. Let `r` the radius of the sphere and a be the linear acceleration of the centre of the sphere the angular acceleration about the centre of the spohere is `alpha=a/r`, as there is no slipping. For the inear motion of the centre `F+f=ma`.............i and for the rotational motion about the centre, `Fr-fr=ialpha=(2/3mr^(2))a/r` or `F-f=2/5ma`.............ii from eqn i andn ii we get `2F=7/5 ma` or `a (10F)/(7m)` Alternative method: As the shre is rolling, we can apply torque equation about point of contact. i.e. `F(2r)=I_(p)alpha=(2/5mr^(2)+mr^(2))alpha` `impliesF(2r)=(7/5mr^(2))alpha` Hence angular acceleration `alphas=(10F)/(7mr)`..........iii As sphere is rolling, acceleration of centre of mass is `a=alphar=(10F)/(7m)`
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