Rolling Is cool!

Classical Mechanics Level 5

A uniform cylinder is given an angular velocity of $\omega = 5\text{ rad/sec}$ in clockwise direction and dropped from the height $h = 22\text { m}$ .

It collides with the sufficiently rough ground such that its component of velocity in the vertical direction reduces to zero. And just after the impact, it starts pure rolling on the ground.

Find the minimum coefficient of friction of the rough ground required $\min( \mu )$ .

Details and Assumptions :

The radius of cylinder is $R = 2\text{ m}$ .
The initial height measured is from the centre of mass of cylinder from ground.
Acceleration due to gravity, $g = 10\text{ m/s}^2$ .

Try more here

The answer is 0.1667.

1 solution

Prakhar Bindal
Apr 14, 2016

Nice problem! :)

Firstly as there is no external torque(moment or couple) of any force about Centre of mass Hence its Angular Velocity will remain constant till the cylinder strikes the Surface.

Using work energy theorem

W = Net change in kinetic energy

Mg(H-R) = 1/2 m v^2 1....

Now carefully Analyze the ball on striking the surface.

There will be Two Impulsive forces acting on it (We can neglect finite forces like gravity when impulsive forces come into play)

Using Impulse Momentum Theorem we have

Call N to be the normal reaction provided by the surface and f be friction.

We Have

Ndt = mv (v is velocity of cylinder just before striking the surface)

Angular Impulse about centre of mass will be only due to friction

fdt*R = Change in Angular Momentum = I (w-5)

I Denotes moment of inertia about centre of mass and w is final angular velocity of cylinder

We also have to do pure rolling V2 = Rw (For bottom point to remain stationary)

V2 Is horizontal velocity of Centre of mass after striking along horizontal surface

Again Using impulse momentum theorem for frictional force in horizontal direction

fdt= MV2

Using all equations and solving

We obtain Coefficient of friction = 1/6

Exactly:) ,,,but u did a mistake fdt * R = I (5 - w).....on considering the directions carefully

Aniket Sanghi - 5 years, 2 months ago

I Assumed friction in opposite direction thats why when i calculated fdt i got it as negative

Prakhar Bindal - 5 years, 2 months ago

Ok!! Then that's right!

Aniket Sanghi - 5 years, 2 months ago

@Aniket Sanghi – No Problem! :)

Prakhar Bindal - 5 years, 2 months ago

Yup!Same method!Easy but nice!P.S:for finding the velocity of the cylinder when it hits the ground you could also have used the class 9 equation $v^2-u^2=2as$ :P

Adarsh Kumar - 5 years, 1 month ago

If you feel it easy try aniket's problem lets play football it will be surely challenging!

Prakhar Bindal - 5 years, 1 month ago

I tried that but couldn't solve it!

Adarsh Kumar - 5 years, 1 month ago

@Adarsh Kumar – Oh!!! Did u like it ??

Aniket Sanghi - 5 years, 1 month ago

@Aniket Sanghi – I had a few doubts regarding the problem and the solution,I have commented there,could you help me out?

Adarsh Kumar - 5 years, 1 month ago

Thanks! it's my own :)

Aniket Sanghi - 5 years, 1 month ago

but bro , ur football problem , i know i can solve it , but won't be it a little lengthy to calculate horizontal , vertical velocities and the changes in angular velocity every time it hits and then calculate until u get a zero horizontal velocity !

A Former Brilliant Member - 4 years, 7 months ago

@A Former Brilliant Member – That's easy ! And I promise that that's not that lengthy . If you are committing mistaking then u will feel it to be lengthy! Check yr method once!

Aniket Sanghi - 4 years, 7 months ago

took it as a hollow cylinder MI=mr^

sashank bonda - 4 years, 8 months ago

wth ! i was typing 0.16 from the beginning and it was saying wrong but at last ..... i closed my eyes , wrote 0.166666666667 and taped enter and whoa it was right ! brilliant should do something about it ! i suffer like this may a times . do't you agree ?? BTW great problem bro Aniket !

A Former Brilliant Member - 4 years, 7 months ago

I agree!

Thanks !

Aniket Sanghi - 4 years, 7 months ago

I did the same mistake(Didn't take care of directions) and got 0.5 as the answer!!

Harry Jones - 3 years, 8 months ago

Interesting problem

Dehia Zaman Sharup - 8 months, 2 weeks ago

i am getting answer 0.5 .... why?i am getting answer as 0.5 .. why?

Dhruv Aggarwal - 5 years, 1 month ago

What was your approach? If you did it the way it's given above then it might be a computation error. I'll just drop the primary equations here that you should get so that you can check your working by yourself.

Notations:

$J=$ Impulse due to Normal Reaction Force.

$v=$ Velocity of the cylinder as it hits the ground.

$v'=$ Velocity with wich it starts pure rolling.

$m=$ Mass of the cylinder.

$I=$ Moment of Inertia of the cylinder.

Equations:

$v^2=2g(h-R)$

$J=mv$ (By Impulse-Momentum Theorem )

$\mu J = mv'$ (By Impulse-Momentum Theorem )

$\mu JR = I\omega-I\large \frac{v'}{R}$ (By Angular Impulse-Angular Momentum Theorem )

Miraj Shah - 5 years, 1 month ago

Can you please explain how did you get the term (coeff. Of friction)*J in eqn 3,4?

Nihar Mahajan - 3 years, 7 months ago

ya may be i was making a mistake earlier got the answer as 1/6...

Anyways thanks for ur help and plz edit ur first eqn.. should be h-r

Dhruv Aggarwal - 5 years, 1 month ago

Rolling Is cool!

Try more here

The answer is 0.1667.

1 solution

0 pending reports