Sliding and Rolling

Consider the following example. A cylinder of mass M and radius R is held at rest on a plane inclined at an angle to    to the horizontal. When the cylinder is released it moves down the slope. Let us discuss the motion of the cylinder.

When the cylinder is released, the component Mg sin    of its weight Mg along the incline causes the cylinder to slide down the slope. If the force of friction between cylinder and incline is ƒ, we get

Mg sin    - ƒ = Ma_c

where, a_c is the linear acceleration of CM of the cylinder.

The frictional force provides a rotational torque about the CM axis of the cylinder:

ƒR = I_c   

where, I_c = Mk² is the moment of inertia of the cylinder about its CM axis (which is a principal axis); k is radius of gyration. The axis of rotation does not change its direction in space, or with respect to cylinder.

with constant acceleration, a_c, the CM velocity increases with time as,

v_c = v₀ + a_c t = a_c t                       (v₀ = 0)

Simultaneously, with constant angular acceleration   , the angular velocity of rotation of the cylinder increases as,

ω = ω₀ +    t =    t                         (ω₀ = 0)
    
Rolling: If a_c =    R, then vP = 0 independent of t. That is, if,

 

then the object performs pure rolling throughout the motion, there is no sliding.

However, the frictional force cannot exceed its limiting value, ƒ_max =    N =    Mg cos   . Hence, so long as ƒ given is less than (or at most equal to) ƒ_max, the frictional force would acquire the value required to support rolling. Thus, the object will perform only pure rolling, if

 

For the cylinder, k² = R²/2. Hence, we find that if    ≥ 1/3 tan    then a cylinder kept on an inclined plane will purely roll down. The acceleration of its CM then is,

 

= 2/3 g sin          (for cylinder)

The rolling is therefore equivalent to translation of CM with acceleration a_c and rotation about CM axis with angular acceleration   = a_c/R. The kinetic energy of rolling object (the cylinder) is, therefore

 

   

where we used the fact that v_c = ω R in rolling, and I_c = Mk².

For rolling cylinder, we have   
    
Rolling with slipping: If    then the value of ƒ is not physically possible. This in turn implies that ac =    R, orv_c = ω R are not possible and therefore the object can never perform pure rolling. The object therefore will always slide and also rotate, as it moves down the slope.

Now, the frictional force is equal to force of sliding friction, i.e.

ƒ =    Mg cos

Both the linear and angular accelerations (and velocities) are given by the same relations as before:

a_c =  (sin    -    cos   ) g, v_c = a_c t

 

The object translates with same linear acceleration as in the case of pure sliding, but also simultaneously rotates.

Posted Date: 9/18/2012 1:59:50 AM | Location : United States

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