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.

Things to Remember

mathematical form of acceration of rolling body on inclined plane.

MCQs

No MCQs found.

Subjective Questions

No subjective questions found.

Videos

No videos found.

Acceleration of an Object Rolling Down an Inclined Plane

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.

Fig: Body rolling on inclined plane
Fig: Body rolling an inclined plane

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

Recent Notes

No recent notes.

Related Notes

No related notes.