The following question on rolling bodies will be worth noting since many among you will be able to correct certain misconceptions:

**A disc made of iron rolling along a horizontal surface with a velocity ‘v’**** encounters an inclined plane of inclination 30****º as shown in the figure. It rolls up the plane (without slipping) and reaches a height ‘h’ before turning back. If a solid sphere made of iron were rolling up with the same initial velocity, the height reached before turning back would be (ignoring the loss of energy against friction)**

**(a) h (b) 7h/8 (c) 9h/10 (d) 12h/13 (e) 14h/15 **

The entire initial kinetic energy of the disc (or sphere) gets converted into gravitational potential energy on reaching the maximum height. Since this is a case of rolling, the kinetic energy is partly translational and partly rotational. In the case of the disc we have

½ Mv^{2}+ ½ Iω^{2} = Mgh, where ‘M’ is the mass, ‘I’ is the moment of inertia, ‘ω’ is the angular velocity and ‘g’ is the acceleration due to gravity.

Since ω = v/R and I = MR^{2}/2, this can be rewritten as

¾ Mv^{2} = Mgh, from which h = 3v^{2}/4g

In the case of the solid sphere The energy equation, ½ Mv^{2}+ ½ Iω^{2} = Mgh can be rewritten as

½ Mv^{2}+ ½ (2/5)Mr^{2}(v^{2}/r^{2}) = Mgh ^{}

We have used the symbol ‘r’ for the radius of the sphere. This simplifies to

(7/10)Mv^{2} = Mgx, where ‘x’ is the maximum height reached by the sphere.

Therefore, x = 7v^{2}/10g = (7/10)(4/3)(3/4)v^{2}/g = **14h/15, **since h = 3v^{2}/4g.

Whenever you solve problems, you should think of other possibilities. A question setter can modify the above question as follows:

**A disc and a solid sphere made of the same material have the same mass but radii in the ratio 3:1. If they have the same initial velocity, the ratio of the maximum heights to which they can roll up an inclined plane of given inclination is (ignoring the work done against friction) **

**(a) 15/14 (b) 14/15 (c) 45/14 (d) 42/15 (e) 3 **

We have seen in the previous problem that the height is independent of the radius. The correct option is 15/14 (since we require the ratio h/x).

You should note that the two bodies need not be of the same material as the work done against friction can be ignored.

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