Acceleration of an Object Rolling Down an Inclined Plane
The greater the value of K, at compared with R, the smaller the acceleration of the body rolling down along the plane and hence greater the time taken by it in reaching the bottom of the plane. Also, the acceleration and hence the time of descent, is quite independent of the mass of the body.
Summary
The greater the value of K, at compared with R, the smaller the acceleration of the body rolling down along the plane and hence greater the time taken by it in reaching the bottom of the plane. Also, the acceleration and hence the time of descent, is quite independent of the mass of the body.
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mathematical form of acceration of rolling body on inclined plane.
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Acceleration of an Object Rolling Down an Inclined Plane
Acceleration of an Object Rolling down an Inclined Plane:
Consider a rigid spherical object of mass M and radius R, which is initially at the top of an inclined plane of height ‘h’ inclined at an angle ‘θ’ with the horizontal.

Initially when the object is at the top of the inclined plane it has only potential energy. But when it is released from the top of the inclined plane free it falls down the plane. Finally when the body reach the horizontal surface it possess only the kinetic energy which is equal to the potential energy of the body on the top of the inclined plane. i.e.
\begin{align*}Mgh=\frac{MV^{2}}{2}(1+\frac{K^{2}}{R^{2}})\end{align*}\begin{align*}V^{2}=\frac{2gh}{(1+\frac{K^{2}}{R^{2}})}--------(1)\end{align*}
Where, K is the radius of gyration of the body and v the velocity of the body at the bottom of the inclined plane.
If ‘S’ be the length of the inclined plane and ‘a’ be the acceleration of the body, then from equation of motion, we can write,
\begin{align*}V^{2}=u^{2}+2as\end{align*}\begin{align*}V^{2}=2as--------(2)(\because u=0)\end{align*}
Equating equation (1) and (2), we get,
\begin{align*}
\frac{2gh}{(1+\frac{K^{2}}{R^{2}})}=2as\end{align*}\begin{align*}a=\frac{gsin\Theta }{(1+\frac{K^{2}}{R^{2}})}\end{align*}\begin{align*}\therefore a=(\frac{R^{2}}{K^{2}+R^{2}})gsin\Theta\end{align*}
Hence, \(\begin{align*}a\alpha (\frac{R^{2}}{K^{2}+R^{2}})\end{align*}\) for a given angle of inclination of the plane.
The greater the value of K, at compared with R, the smaller the acceleration of the body rolling down along the plane and hence greater the time taken by it in reaching the bottom of the plane. Also, the acceleration and hence the time of descent, is quite independent of the mass of the body.
References:
Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.
Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.
Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.
Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014
Lesson
Dynamics of Rigid Bodies
Subject
Physics
Grade
Bachelor of Science
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