If a sphere is rolling, the ratio of its rotational K.E to total energy is given by

A solid sphere is moving on a horizontal plane. Ratio of its translational K.E. and rotational kinetic energy is [Hint : E_L/E_r = (1/2mv^2)/(1/2Iomega^2) = (mv^2)/(2/5mr^2 omega^2) = (mv^2)/(2/5mv^2) = 5/2 ]

A partical of mass m1 is travelling with avelocity u1 strikes head-on against partical of mass m2 at rest.After collision the particales move with a velocity v1 and v2 respectively .show that the ratio of decrese in K.E of the partical of mass m1 to its original K.E is given By (4m1m2)/(m1+m2)^2

A spherical ball rolls on a table without slipping .The fraction of its total energy associated with rotations

A solid cylinder rolles down an inclined plane. Its mass is 2 kg and radius 0.1 m. If the height of the inclined plane is 4m, what is its rotational kinetic energy when it reaches the bottom.

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v^2=(2gh)/ (i+k^2^l R^3 Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

When a body falls freely towards the earth, then its total energy-

Find the ratio of total kinetic energy to the rotational kinetic energy of a rolling sphere moving along a horizontal plane.

Two spheres made of same meterial have radii in the ratio 1:2 both are the same temperature Ratio of heat radiation energy emitted per sec by them is

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR^2IS, where M is the mass of the sphere and R is the radius of the sphere(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2I4, find its moment of inertia about an axis normal to the' disc and passing through a point on its edge.

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx^2/2, where k is the force constant of the oscillator. For k = 0.5 N ^-1 , the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.,br>