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What is the relation between P.E and K.E of cylinder for rolling motion?

What is the relation between P.E and K.E of cylinder for rolling motion?

Grade:12th pass

1 Answers

Shalini
16 Points
6 years ago
In experiment, we will apply the Law of Conservation of Energy to objects rolling down a ramp. As an object rolls down the incline, its gravitational potential energy is converted into both translational and rotational kinetic energy. The translational kinetic energy is
( 1 )
KEtrans = (1/2)mv2 
whereas the rotational kinetic energy is
( 2 )
KErot = (1/2)Iω2 
In this last equation ω is the angular velocity in radians/sec, and I is the object's moment of inertia. For objects with simple circular symmetry (e.g. spheres and cylinders) about the rotational axis, I may be written in the form:
( 3 )
I = kmr2 
where m is the mass of the object and r is its radius. The geometric factor k is a constant which depends on the shape of the object:
  •  
k = 2/5 = 0.4
    
for a uniform solid sphere,
k = 1/2 = 0.5
    
for a uniform disk or solid cylinder,
k = 1
    
for a hoop or hollow cylinder.
If the object rolls without slipping, then the object's linear velocity and angular speed are related by
v = rω. 
Substituting eq 3 and the expression for v into eq 2 we obtain:
( 4 )
KErot = (1/2)kmv2 
 
Consider a round object rolling down a ramp as in the illustration above. Assuming no loss of energy we may write the conservation of energy equation as:
               total energy at top of ramp = total energy at bottom of ramp,
               Egravitational = Etranslational + Erotational
 
or,
( 5 )
mgh = (1/2)mv2 + (1/2)kmv2. 

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