Sliding and rolling, Mechanical Engineering

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  817_download.png  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  817_download.png  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  817_download.png  - ƒ = Mac

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

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

ƒR = Ic  817_download.png 

where, Ic = 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, ac, the CM velocity increases with time as,

vc = v0 + ac t = ac t                       (v0 = 0)

Simultaneously, with constant angular acceleration  817_download.png , the angular velocity of rotation of the cylinder increases as,

ω = ω0 +  817_download.png  t =  817_download.png  t                         (ω0 = 0)
    
Rolling: If a=  817_download.png  R, then vP = 0 independent of t. That is, if,

1984_download (1).png 

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

However, the frictional force cannot exceed its limiting value, ƒmax =  928_download (2).png  N =  928_download (2).png  Mg cos  817_download.png . Hence, so long as ƒ given is less than (or at most equal to) ƒmax, the frictional force would acquire the value required to support rolling. Thus, the object will perform only pure rolling, if

1515_download (3).png 

For the cylinder, k2 = R2/2. Hence, we find that if  928_download (2).png  ≥ 1/3 tan  817_download.png  then a cylinder kept on an inclined plane will purely roll down. The acceleration of its CM then is,

451_download (4).png 

= 2/3 g sin  817_download.png        (for cylinder)

The rolling is therefore equivalent to translation of CM with acceleration ac and rotation about CM axis with angular acceleration  817_download.png = ac/R. The kinetic energy of rolling object (the cylinder) is, therefore

2386_download (5).png 

  2485_download (9).png 

where we used the fact that vc = ω R in rolling, and Ic = Mk2.

For rolling cylinder, we have  312_download (6).png 
    
Rolling with slipping: If  1299_download (7).png  then the value of ƒ is not physically possible. This in turn implies that ac =  817_download.png  R, orvc = 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 =  928_download (2).png  Mg cos

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

ac =  (sin  817_download.png  -  817_download.png  cos  817_download.png ) g, vc = ac t

426_download (8).png 

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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