Rotational Motion – Rolling bodies

If you release from rest differently shaped regular bodies such as disk, ring, hollow sphere, solid sphere, hollow cylinder(pipe) and solid cylinder from the top of an inclined plane, thereby allowing them to roll down the plane, you will find that the solid sphere always arrives at the bottom first and the ring and the pipe arrive last. The acceleration ‘a’ of a body of radius ‘R’ rolling down an inclined plane (without slipping) is given by a= g sinθ/[1+( k² / R²)] where ‘g’ is the acceleration due to gravity, ‘θ’ is the angle of the plane and ‘k’ is the radius of gyration given by I= Mk² where ‘I’ is the moment of inertia of the body (of mass M) about the central axis about which it is rolling down. Since the moment of inertia of a solid sphere about its diameter is (2/5) MR², we have (2/5)MR² = Mk². Therefore, k²/R² = 2/5.
The ring and the pipe have k²/R² = 1, since I=MR² = Mk². If you find the values of k²/R² in the case of differently shaped rolling bodies, you will realize that it is minimum (2/5) for the solid sphere and maximum (equal to1) for a thin ring and a thin pipe. Since (k²/R²) appears in the denominator in the expression for ‘a’, the acceleration down the plane is maximum in the case of a solid sphere and minimum in the case of a ring or a pipe. So the solid sphere arrives at the bottom first and the ring and the pipe arrive last.
You should remember that the moment of inertia of a disk (and that of a solid cylinder) about its rolling axis is ½ MR² and that of a hollow sphere is (2/3) MR²
It is interesting to note that all solid spheres will arrive at the bottom together, irrespective of their mass and size. Similarly, all thin rings and thin pipes will arrive together, irrespective of their mass and size. Generally speaking, bodies of a given shape will arrive together, irrespective of their mass and size.
Now, consider the following M.C.Q.:

A body of mass M at the top of a smooth inclined plane starts from rest and slides down the plane. It reaches the bottom with a velocity ‘v’. If the same body were in the form of a ring and the plane were rough, the velocity with which the ring will reach the bottom on rolling down the plane would be
(a) v (b) v√2 (c) 2v (d) v/√2 (e) v/2
Note that if a body is to roll along any surface, friction is necessary and that is why the plane is said to be made rough in the problem. On a smooth plane, the body will slide down with acceleration gsinθ and will reach the bottom with a velocity ‘v’ given by v² = 0 + 2gsinθ×s, as given by the usual equation of linear motion, v² = u² + 2as.
Therefore, v = √(2sgsinθ).
On rolling, the acceleration down the plane is gsinθ/[1 + (k²/R²)] = ½ gsinθ since k²/R² = 1 for a ring. Therefore the velocity (v1) on reaching the bottom is given by v₁² = 0 + 2×½ gsinθ×s
Therefore, v₁ = √(sgsinθ) = v/√2 [Option (d)].

Let us discuss another M.C.Q.:

A sphere of radius ‘R’ is kept at the top end of a curved track. The upper end of the track is at a height ‘H’ and the lower end which is horizontal, is at a height ‘h’ above above the ground level.

When the sphere is released, it rolls down (without slipping) along the track and after leaving the lower end of the track, it moves like a projectile and lands at the point C. The horizontal distance BC is
(a) R√[(20/7)gh(H-h)] (b) √[(20/7)gh(H-h)] (c) √[(10/7)gh(H-h)]
(d) √[(20/7)h(H-h)] (e) √[(5/7)h(H-h)]
The loss of potential energy by the sphere on rolling down along the track is equal to the gain of kinetic energy (both translational and rotational) so that
Mg(H-h) = ½ Mv² + ½ I ω² where M is the mass, ‘v’ is the linear velocity, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the sphere, which is (2/5)MR². Substituting for ω = v/R, we obtain v = √[(10/7)g(H-h)]. This is the horizontal velocity of the sphere at the lower end of the track. Since its height at the lower end of the track is ‘h’, the time taken to reach the ground from there is t = √(2h/g). Note that we obtain this time by considering the vertical motion of the sphere and using the equation, h = 0 + ½ gt². The horizontal distance BC = vt = √[(20/7)h(H-h)]. So, (d) is the correct option. Note that this contains neither R nor g.
The following question is a conventional type which you will get in your class examinations as well as in entrance tests:

A solid cylinder of mass M and radius R rolls down an inclined plane of height ‘h’ without slipping after starting from rest at the top. The speed of its centre of mass when it reaches the bottom is
(a) √(2gh) (b) √(4gh/3) (c) √(3gh/4) (d) √(2gh/3) (e) √(3gh/2)
Equating the gravitational potential energy of the cylinder at the top of the plane to the sum of the translational and rotational kinetic energies at the bottom, we have,Mgh = ½ Mv² + ½ I ω² = ½ Mv² + ½ ×½ MR² × (v/R)² = (3/4)Mv², from which v = √(4gh/3).

Here is another typical M.C.Q. involving rotational motion:
A spherical ball rolls on a table without slipping. The fraction of its total energy which is associated with rotational motion is
(a) 3/5 9b) 2/3 (c) 2/5 (d) 3/7 (e) 2/7
The rolling ball has translational and rotational kinetic energies giving it total kinetic energy equal to ½ Mv² + ½ I ω² = ½ Mv² + ½ ×(2/5)MR² × (v/R))² =½ Mv² + (1/5)Mv² = (7/10)Mv².
The rotational kinetic energy is (1/5)Mv² and the total kinetic energy is (7/10)Mv². The required ratio is therefore (1/5) / (7/10) which is 2/7 [Option (e)].

You will see interesting questions (with solution) on rotational motion here as well as here.

If you would like to have still more multiple choice questions (with solution) on rotational motion, you can have them here.