# 6. Physics | Simple Harmonic Motion | A Uniform Rod on Two Rollers | by Ashish Arora

in this illustration we’ll discuss about a

uniform rod on 2 rollers. here we are given that a uniform rod is placed on 2 spinning

rollers symmetrically as shown in figure. and the friction coefficient between the rod

and rollers is mu. and we are required to find the time period of, small oscillations

of rod if it is slightly displaced horizontally and released. here first we need to understand

this situation we are given 2 rollers. on which a uniform rod is placed. here we can

say if, the rollers are spinning in clockwise and anticlockwise manner. so, the direction

in which friction would be acting on rollers would be opposite to, their spinning direction.

and on rod it will be in the direction opposite to the friction acting on the rollers. initially

when, the weight is acting at its central point, the normal reaction at both of these

rollers would be equal. say these are n and n, we can say the value of n will be, m g

by 2 because, 2 n will be equals to m g for the equilibrium of rod in vertical direction.

and when we slightly displaced the rod toward right say by a distance x we can say. the

center will shift by a distance x and now this is the point where m g would be acting.

this was, the initial, position where m g was acting, and this is after distance x,

when the rod is shifted, m g would be acting here. due to which this normal reaction will

become n 1 and this will become n 2, where n 1 would be certainly greater then n 2 because

m g is shifted toward the rightward roller. so in this situation the friction will also

be this is f 1 and this is f 2, which are acting. on the rod. so here we can first calculate

n 1 and n 2 and in turn we can get the value of friction and difference of the 2 will be,

the restoring force on rod which will tended back to the initial position. so here we can

write, for equilibrium, of rod. after displacing it. by x we use. say if this point is ay,

and this point is b here we can write we use torque about point ay, here we can write about

point ay torque is due to n 1. which is n 1 l, minus it is due to the weight which is

m g multiplied by l by 2 plus x. and this should be equal to zero because the rod is

in equilibrium, for its rotational motion. so this implies here we get the value of n

1 directly as, m g by l multiplied by l by 2 plus x, similarly we can write the torque

about point b, where the torque due to n 1 become zero and it is only due to n 2 and

its weight. so we can write it as, n 2 l minus, m g multiplied by l by 2 minus x because distance

of m g from this point b is l by 2 minus x. this distance is l by 2. this is also l by

2 as initially the weight would be acting at its mid point. this should also be zero.

which gives us the value of n 2. that is m g by l multiplied by l by 2 minus x. so we

can write the restoring force. on rod is. this restoring force we can write this as

f 1 minus f 2. which is mu times, n 1 minus mu times n 2. if we substitute the values

of n 1 and n 2 over here this restoring force becomes. mu times m g by l multiplied by,

l by 2 plus x minus l by 2 again this plus x, so this l by 2 gets cancelled out. and

this will result as m ay if ay is the acceleration of rod. with which it is having a tendency

to return back. so this gives us acceleration is equal to here we can cancel out m. and

this will be 2 mu g by, l multiplied by x. and as ay is opposite to, x we can write as,

ay is in. direction. opposite to x this implies we can write ay as minus 2 mu g by l multiplied

by x. or we can compare it with ay is equal to minus omega square x. so here we can write

it comparing with. ay minus omega square x because here acceleration is directly proportional

to x this implies, it is executing s h m. this implies rod, is executing. s h m at.

omega is equal to root of 2 mu g by, l. this implies its time period of oscillation is

2 pie by omega which is 2 pie, root of l by 2 mu g that will be the final result of this

problem.

## 8 Replies to “6. Physics | Simple Harmonic Motion | A Uniform Rod on Two Rollers | by Ashish Arora”

A ring of mass m can be hinged at a point on its periphery. There are two ways in which it can be made to oscillate : (a) It oscillates in a vertical plane (b) it oscillates in a direction perpendicular to its plane. The time period of oscillation in case (b) is √3 sec . Find the time period of oscillation in case (a).

sir here the torque about point a should be Nl+mg(l/2+x) since they both are acting in clockwise direction.please clear my doubt

sir rather we can balance the torque about COM by writing the equations N1(l/2-x)=N2(l/2+x) and N1+N2=mg in addition to f1-f2=ma. by using these equations i am getting the same answer

is energy conservation applicable here ?

sir thoda hindi bhi use kare but you are awesome

Comprehensive Explaination

Amazing problem!

sir can u explain the direction of f2 (friction 2) . why it would be inward as friction acts opp to relative motion so do i need to assume that at point of contact of rod and roller ,roller would have low tangential velocity?