Question

!!Open Challenge!!

For Mods, Brainly newton's, Geniuses!!

Class=> 11

Topic▶ System of particles and rotational motion.

Hello!

$\sf\huge{\underline{\mathbb{Question}}}$

▶A solid disc and a ring, both of radius 10cm are placed on horizontal table simultaneously, with initial angular speed equal to 10π rad\s. Which of the two will start to roll earlier?? The coeffecient of kinetic friction u(k) = 0.2


With proof!!


▶Hint!

Use the concept of Linear velocity!!(Newton's motion equation)!!


​

Answer 1

Radius of disc and ring (R) = 10cm = 0.1 m

initial angular speed (wo) = 10π rad/s

Coefficient of Kinetic friction (u) = 0.2

Moment of inertia of disc (Id)= mR²/2

Moment of inertia of ring (Ir)= mR²

We know, frictional force responsible for the motion of body .

So, F = fr = ma ---------(1)

Where a is linear acceleration and N is normal reaction .

umg = ma

a = ug ----------(2)

Let ā is the angular acceleration.

We know,

Torque = Iā

Also torque = R×F use this here,

R×F = Iā

Rumg = Iā

ā = umgR/I

For disc

___________

ā = umgR/mR²/2 = 2ug/R

For ring

___________

ā = umgR/mR² = ug/R

Now,use linear equation,

V = u + at

Initial velocity (u) = 0

V = at

V = ugt { from equation (2)

For rolling motion ,

V = wR

w = v/R = ugt/R -------(4)

Let t1 time taken by ring and t2 time taken by disc .

use , angular equation ,

w = wo + āt

For disc

___________

w = wo + āt2

ugt2/R = wo -2ugt2/R { from equation (4) , ā = -2ug/R frictional torque oppses the motion of the body }

t2 = Rwo/3ug

Put R , wo , u and g values

t2 = 0.1 × 10π/3×0.2 ×9.8 = 0.53 s

For ring

___,_________

w = wo + āt1

ugt1/R = wo - ugt1/R

t1 = Rwo/2ug

Put R, wo , u and g values

t1 = 0.1 × 10π/2×0.2×9.8 =0.8 s

It is clear that disc will start rolling earlier .

Answer 2

Answer:

Explanation:

Radius of disc and ring (R) = 10cm = 0.1 m

initial angular speed (wo) = 10π rad/s

Coefficient of Kinetic friction (u) = 0.2

Moment of inertia of disc (Id)= mR²/2

Moment of inertia of ring (Ir)= mR²

We know, frictional force responsible for the motion of body .

So, F = fr = ma ---------(1)

Where a is linear acceleration and N is normal reaction .

umg = ma

a = ug ----------(2)

Let ā is the angular acceleration.

We know,

Torque = Iā

Also torque = R×F use this here,

R×F = Iā

Rumg = Iā

ā = umgR/I

For disc

___________

ā = umgR/mR²/2 = 2ug/R

For ring

___________

ā = umgR/mR² = ug/R

Now,use linear equation,

V = u + at

Initial velocity (u) = 0

V = at

V = ugt { from equation (2)

For rolling motion ,

V = wR

w = v/R = ugt/R -------(4)

Let t1 time taken by ring and t2 time taken by disc .

use , angular equation ,

w = wo + āt

For disc

___________

w = wo + āt2

ugt2/R = wo -2ugt2/R { from equation (4) , ā = -2ug/R frictional torque oppses the motion of the body }

t2 = Rwo/3ug

Put R , wo , u and g values

t2 = 0.1 × 10π/3×0.2 ×9.8 = 0.53 s

For ring

___,_________

w = wo + āt1

ugt1/R = wo - ugt1/R

t1 = Rwo/2ug

Put R, wo , u and g values

t1 = 0.1 × 10π/2×0.2×9.8 =0.8 s

It is clear that disc will start rolling earlier .