A particle P, free to move on the smooth inner surface of a fixed hollow sphere of centre O and radius a is placed at the lowest point of A of this surface. It is then projected horizontally with...

A particle P, free to move on the smooth inner surface of a fixed hollow sphere of centre O and radius a is placed at the lowest point of A of this surface. It is then projected horizontally with initial speed `sqrt(nga)` where n>0.

While particle is still in contact with the surface find the reaction on the particle from the surface when the angle turned by OP is `theta`

 

2 Answers | Add Yours

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

We can use energy conservation law here assuming that there is no energy losses during the motion.

Let us define the horizontal line that passes through A as datum and the mass of P as m.

When P is a A;

Energy `(E) = 1/2xxmxx(nga)-mga`

 

When particle is at an theta angle;

`E = 1/2xxmxxv^2-mg(a-acostheta)`

 

`1/2xxmxx(nga) = 1/2xxmxxv^2+mg(a-acostheta)`

`nga = v^2+2ga(1-costheta)`

`v^2 = ga(n-2+2costheta)`

 

Using Newtons second law when P is at an angle `theta` towards direction of the centre of the circle which is OP;

`F = ma`

`R-mgcostheta = mv^2/a`

`R = mgcostheta+m(ga(n-2+2costheta))/a`

`R = mg(n-2+3costheta)`

 

So reaction(R) on the particle from the surface when the angle turned by OP is theta is `mg(n-2+3costheta)`.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Sources:
llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The particle, of mass m, is projected from the lowest point A of a sphere with velocity `u=sqrt(nga)` to move along the inside of the hollow sphere (refer to the attached image).

Let P be the position of the particle at any time t where arc AP = s and angle AOP = `theta` . If v be the velocity of the particle at P, the equations of motion along the tangent and normal are:

`m*(d^2s)/dt^2=-mgsintheta` ..........(i)

and `(mv^2)/a=(R-mgcostheta)` ..........(ii)

also, `s=a*theta` ........(iii)

From (i) and (iii),

`a*(d^2theta)/(dt^2)=-gsintheta`

Multiplying both sides by `2a(d theta)/(dt)` and integrating,

`2a^2*int((d^2theta)/(dt^2)*(d theta)/(dt))=-2agint(sintheta*(d theta)/(dt))` [for LHS integral, put `x=(d theta)/dt, dx=(d^2theta)/(dt^2)` and it reduces to `int(xdx)=1/2 x^2` ]

So, `(a(d theta)/(dt))^2=2ag*costheta+C`

`rArr v^2=2ag*costheta+C` (since `v=a*(d theta)/(dt)` )

To evaluate C, the integration constant,

At A, when  `theta=0, v=u=sqrt(nga)`

`u^2=2ag*1+c`

`rArr c=(u^2-2ag)`

So, `v^2=2agcostheta+ u^2-2ag`

`=2agcostheta+ nag-2ag`

`=ag(2costheta+ n-2)`

Hence the normal reaction, from eqn. (ii) we have

`R=m/a(v^2+agcostheta)`

`= m/a(ag(2costheta+ n-2)+agcostheta)`

`=mg(3costheta+n-2)`

Therefore, the normal reaction of the surface on the particle while it is still in contact with the surface is given by

`R= mg(3costheta+n-2)`

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Sources:

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question