Rolling Motion on an Inclined Plane
It is an example of the rigid body motion, in which the centre of mass is also in motion.
Consider a solid object (sphere) of radius R and mass m rolling down a plane inclined at an angle 0 to the horizontal figure. The height of the inclined plane is h and the distance is s. If the object rolls down from the highest point of the inclined plane then at that time the total energy at point C = potential energy
= mgh …………. (1)
In this motion the object does the rotational motion about an axis passing through the center of mass as well as the translatory motion along the plane.
Thus, Rolling motion = Translatory motion + Rotatory motion.
At the bottom of the plane A, its velocity becomes v and angular velocity ω, then
Total energy at point A= Translatory kinetic energy + rotational energy
= \(\frac{1}{2}\) mv2 + \(\frac{1}{2}\) Iω2
Hence, by the law of conservation momentum
This is the representation of the velocity of a body at an inclined plane.
If the initial velocity of the body is zero (u = 0) then by the equation of motion (III rd equation)
This is the formula for acceleration of a body on an inclined plane.
If the objects of various shapes are rolled down an inclined plane then the object whose K2 / R2 is minimum, its value of velocity and acceleration will be maximum when it reaches the end; and it will reach first at the bottom; the object whose K2 / R2 value is maximum will take maximum time to reach the end.
The value of velocity, acceleration and time are not depend upon the mass of a rolling body on an inclined plane. For similar value of h, the value of velocity is not depend upon the angle ‘θ’ of the inclined plane.
The value of K2 / R2 for main objects rolling down an inclined plane are as follows :
For solid sphere,
I = MK2 = \(\frac{2}{5}\) MR2 ⇒ \(\frac{K^{2}}{R^{2}}=\frac{2}{5}\)
If all the objects are dropped down together on an inclined plane, then sphere will reach first and ring will be last. Wheel and solid cylinder will reach together at the end.
