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Rolling Motion
KINETIC ENERGY OF ROLLING MOTION
Consider a body of mass M and radius R rolling on a level surface witho slipping. During rolling, suppose that the body rotates about its symmetry axis wi the angular velocity \(\omega\) and its centre of mass moves with the velocity \( v_{\text{C}\text{. M}\text{.}} \) The kinetic energy due to the translational motion of the body,
K. E. (trans) =\( \frac{1} {2}Mv_{\text{C}\text{. M}\text{.}} ^2 \)
and kinetic energy due to the rotational motion of the body,
K. E. (rot) = \( \frac{1} {2}I\omega ^2 \)
where I is moment of inertia of the body about its symmetry axis.
Therefore, kinetic energy of the body during its rolling motion,
K. E. (rolling)= K. E. (trans)+ K. E. (rot)
=\( \frac{1} {2}Mv_{\text{C}\text{. M}\text{.}} ^2 \)+\( \frac{1} {2}I\omega ^2 \)
If K is radius of gyration of the body about its symmetry axis, then
I= \( MK^2 \)
\( \therefore \text{K}\text{. E}\text{. (rolling)} = \frac{1} {2}Mv_{\text{C}\text{. M}\text{.}} ^2 + \frac{1} {2}MK^2 \left( {\frac{{v_{\text{C}\text{. M}\text{.}} }} {R}} \right)^2 \)
OR \( \text{K}\text{. E}\text{. (rolling)} = \frac{1} {2}Mv_{\text{C}\text{. M}\text{.}} ^2 + \left( {1 + \frac{{K^2 }} {{R^2 }}} \right) \) -----------(a)
The equation (a) gives the kinetic energy of a body during its rolling moti general. By setting the value of radius of gyration K, the kinetic energy of the b can be found.
It is an example of the rigid body motion, in which the centre of mass is in
motion.
a.) Fraction of translational KE in the total K.E is
\( \frac{1} {{\left( {1 + \frac{{K^2 }} {{R^2 }}} \right)}} = \frac{{R^2 }} {{R^2 + K^2 }} \)
b) Fraction of rotational KE in the total K.E is
\( \frac{1} {{\left( {1 + \frac{{R^2 }} {{K^2 }}} \right)}} = \frac{{K^2 }} {{R^2 + K^2 }} \)
c) Ratio between translational and rotational K.E is \( \frac{{R^2 }} {{K^2 }} \) -
Rolling Motion
ROLLING MOTION OF A BODY (CYLINDER )ON INCLINED SURFACE:
Consider a body( solid cylinder of radius R and mass M) rolling down a inclined at an angle\(\theta\) to the horizontal .
The rolling cylinder is acted upon by the following forces:
(1) Weight Mg of the cylinder acts vertically downwards.
(ii) The friction F between the cylinder and the surface of the inclined pla opposite to the direction of motion.
The weight of the cylinder can be resolved into two components:
(1) Mg sin\(\theta\) along the inclined plane and in downward direction. It ma body to move downwards.
(ii) Mg cos\(\theta\) along perpendicular to the inclined plane. The inclined plan equal and opposite normal reaction ’N' to the cylinder.
N = Mg cos\(\theta\)
Let a be the linear acceleration of the cylinder and F be the limiting: between the cylinder and the surface of the inclined plane. Then, accor Newton's second law of motion
Ma=Mg sin\(\theta\)-F..........(1)
The cylinder rolling down the inclined plane rotates about its symmetry axis .If moment of inertia and a the angular acceleration about the axis of rotation.Then torque on the cylinder is given by
r=I \(\alpha\) ..........(2)
Now, weight of the cylinder and the normal reaction do not contribut orque on the cylinder. It is because both are radial forces. The contribution torque comes from the tangential force only, which is friction. Therefore,
T = FR..........(3)
From the equations (2) and (3), we have
FR=I\(\alpha\)
or F= \( \frac{{I\alpha }} {R} = \frac{I} {R} \times \frac{a} {R} \) \( \left( {\because a = \frac{a} {R}} \right) \)
or F \( = \frac{I} {{R^2 }}a \) ........(4)
In the equation (1), substituting for F, we have
Ma=Mgsin\( \theta - \frac{I} {{R^2 }}a \)
or \( \left( {M + \frac{I} {{R^2 }}} \right)a = Mg\sin \theta \)
or a=\( \frac{{Mg\sin \theta }} {{M + \frac{I} {{R^2 }}}} \).............(5)
In place of moment of inertia, the above relation can be expresses in terms of radius of gyration. If K is radius of gyration of the body about its symmetry axis,then
I= \( MK^2 \)
therefore, the equation (5) gives
a= \( \frac{{Mg\sin \theta }} {{M + \frac{{MK^2 }} {{R^2 }}}}\)
or a=\( \frac{{g\sin \theta }} {{\left( {1 + \frac{{K^2 }} {{R^2 }}} \right)}} \)............(6)
The equations (5) and (6) give the acceleration of a body (capable of rolling) in general, down an inclined plane of inclination \(\theta\). By setting the value of moment of inertia I or radius of gyration K, the acceleration of the body can be found. -
Rolling Motion
Acceleration of a cylinder rolling down the inclined plane.
The moment of inertia of a cylinder about its symmetry axis,
I= \( \frac{1} {2}MR^2 \)
In the equation (5), substituting for I, we have
a= \( = \frac{{Mg\sin \theta }} {{M + \frac{1} {2}\frac{{MR^2 }} {{R^2 }}}} \)
or a=\( \frac{2} {3}g\sin \theta \)............(7)
Condition for the rolling motion of the cylinder. In the equation (4), substituting for I and a, we have
\( F = \frac{1} {2}MR^2 \times \frac{1} {{R^2 }} \times \frac{2} {3}g\sin \theta \)
or F= \( \frac{1} {3}Mg\sin \theta \)..........(8)
If \(\mu_s\) is coefficient of static friction, then
\( \mu _s = \frac{F} {R} = \frac{{\frac{1} {3}Mg\sin \theta }} {{Mg\cos \theta }} \)
or \( \mu _s = \frac{1} {3}\tan \theta \) .........(9)
The equation (9) is the condition for the rolling motion of the cylinder along the inclined plane without slipping.
Note. The condition for rolling motion of a body, in general, can be obtained by setting the value of ‘a’ in the equation (4) as detailed below:
\( F = \frac{I} {{R^2 }} \times \frac{{Mg\sin \theta }} {{M + \frac{I} {{R^2 }}}}\)
\( F = \frac{{Mg\sin \theta }} {{\frac{{MR^2 }} {{I + 1}}}} \) ........(10) -
Rolling Motion
Kinetic energy Comparison among various rolling bodies
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Rolling Motion
Comparison between sliding and rolling on an inclined plane
