Question

ATTRITION: There is normally only one correct snswer for each question snd each correct anwwer ㎞ equal 10 1 polet. Only ehe an" on -answer sheet m will be raluated. Please be sure that you have marked all or your answers on the awer sheet him by ata.Hoa (net pen). Questions 1-2 A ball of mass m and radius r (momeut of inertia Imm) is placed on the inside of A a frictionless circular track of radius Ro as shown in the igure. It starts from rest at thehi vertical edge of the track, and since there is no friction, it slides down without rotation. om fter track starting fronEl isas urace th coefficient of kinetic friction . If the ball starts to roll without slipping after traveling a distance d, what is the expression for the coefficient of kinetic friction in terms of the give

Answer 1

Given,

m is mass of the ball

g is acceleration dueto gravity

R₀ is radius of the track

r is radius of ball

I_cm is moment of inertia about the center of mass =2/5 mr²

To find the relation between coefficient of friction with the parameters given.

We first need to find the velocity of the ball when it reaches point B

We can use the conservation of mechanical energy principle to find it

So,

$U_{i}+K_{i}=U_{f}+K_{f}$

where U is potential energy and K is kinetic energy

From the given figure let us assume the potential energy at the ground is zero

So,

U_i=mgR₀ , the vertical distance of the center of mass at which the ball is placed on the track

K_i=0 as the ball starts from rest so, initial velocity is zero

U_f=mgr as we assumed the potential energy at ground is zero and r is the distance of the ground from center of mass

K_f= 1/2 m v² where v is the speed of the ball when it reaches the ground.

So, putting these values in the energy equation we get,

$mgR_{o}+0=mgr+\frac{1}{2}mv^{2}$

$\Rightarrow v=\sqrt{2g(R_{o}-r)}$

Now, from point B the ball experience friction till its start rolling after a distance d

Since the ball is experiencing friction force it must start slowing down.

we know frictional force is equal to coefficient of friction * normal force by the ground on the object

here, normal force is equal to the weight of the ball = mg

So,

$f_{s}=\mu mg$

and friction is the only force acting along the direction of the horizontal so, the ball has one constant opposing force it must have a constant deceleration

So, acceleration of the object = force applied on the object / mass of the object.

$a=f_{s}/m$

$\Rightarrow a=-\mu mg/m$

$\Rightarrow a=-\mu g$

The negative sign is added as the acceleration is against the motion of the ball.

now, since the deceleration is constant it follows the equations of linear motion

$V^{2}-U^{2}=2AS$

$V-U=At$

$S=Ut+\frac{1}{2}At^{2}$

So, initial velocity is v and final velocity when the ball starts rolling be v_f

$v_{f}^{2}-v^{2}=2(-\mu g)d$............................................(i)

$v_{f}=v-\mu gt$.................................................(ii)

Now,

now, from figure we can see that all forces are acting along the center of mass except friction.

So, the frictional force is solely responsible for the rotational motion that the body starts to experience after reaching point B

Torque due to friction about the center of mass should give us the angular acceleration of the ball

$\tau _{cm}= f_{s}r$

let $\alpha$ be the angular acceleration

So,

$\tau _{cm}= I_{cm}\alpha$

$\Rightarrow f_{s}r=\frac{2}{5}mr^{2}\alpha$

$\Rightarrow \mu mgr=\frac{2}{5}mr^{2}\alpha$

$\Rightarrow \alpha =\frac{5\mu g}{2r}$

now, since the angular acceleration is also constant it follows the same rotational equations of motion as mentiones above.

let  $\omega$ be the final angular velocity when the body starts to roll

initial angular velocity is zero.

$\Rightarrow \omega =0+\alpha t$

$\Rightarrow \omega =\frac{5\mu g}{2r} t$
$\Rightarrow t=\frac{2r\omega }{5\mu g}$

we know $v_{f}=r\omega$ as the ball starts rolling at this instant

$\Rightarrow t=\frac{2v_{f} }{5\mu g}$

putting this value in equation (ii) we get

$v_{f}=v-\mu g*\frac{2v_{f} }{5\mu g}$

$\Rightarrow v_{f}=v-\frac{2v_{f} }{5}$

$\Rightarrow v_{f}+\frac{2v_{f} }{5}=v$

$\Rightarrow\frac{7v_{f} }{5}=v$

$\therefore v_{f}= \frac{5v }{7}$

again putting the value of v_f on equation (i) we get

$\frac{25v}{49}^{2}-v^{2}=2(-\mu g)d$

$\Rightarrow - \frac{24v}{49}^{2}=2(-\mu g)d$

$\Rightarrow d=\frac{24v^{2}}{49*2(\mu g)}$

$\Rightarrow d=\frac{12v^{2}}{49\mu g}$

$\Rightarrow d=\frac{12*2g(R_{o}-r)}{49\mu g}$

$\therefore d=\frac{24g(R_{o}-r)}{49\mu g}$ is the required solution

if any doubt feel free to comment.