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`        What is the relation between P.E and K.E of cylinder for rolling motion?`
2 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 byv = 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.
```
2 years ago
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