Electrostats + Mechanics = Fun

Classical Mechanics Level 5

You took a Hollow spherical ball and gave it an angular velocity $\omega = 10\text{ rad/sec}$ (clockwise) and dropped it from a height $h = 45.5\text{ m}$ . Above the rough horizontal ground ,In a region of uniform Electric field , $E = 6 × {10}^{6}\text{ N/C}$ directed in the horizontal direction (Right).

We observed that after it's collision with the ground it goes to a maximum height of $2.3\text{ m}$ .

Find the Range of the projectile travelled by the ball's centre of mass in between first and second collision.

Details and Assumptions :

The ball has a mass $M = 2\text{ kg}$ and is uniformly charged $( q = 3\mu C )$ . Radius of ball $R = 0.5 \text{ m}$
The height measured is of centre of mass of ball from the ground.
The ball remained in contact with the ground for $t = {10}^{-6} \text{ sec}$ .
Coefficient of friction $\mu = \dfrac{1}{3}$ , take $g = 10\text{ m/s}^2$ .

This is a part of my set Aniket's Mechanics Challenges

The answer is 28.32.

1 solution

Prakhar Bindal
May 6, 2016

Its More of mechanics than of electrostatics but still a very nice problem created by bro @Aniket Sanghi

Notice that the angular velocity initially given to the ball won't change as no net external torque about the centre is there . (Note that electric force do not create any net torque on the ball)

Horizontal force on the ball = qE = 18 N

Horizontal acceleration = 9 m/sec^2

Vertical acceleration = g = 10 m/sec^2

Distance Covered by ball from top to bottom = H-R = 45 m

Vertical velocity attained = 30 m/sec (using standard equation of kinematics)

Horizontal velocity attained = 27 m/sec (Using standard equation of Kinematics)

Consider the moment of striking the surface

We Will Use Impulse Momentum Theorem to solve the problem further

As we can see electric force and gravity are finite forces hence we can neglect them in front of the impulsive normal reaction and frictionn.

Call Ndt impulse due to normal reaction and fdt due friction from ground.

Let After collision ball attains a horizontal velocity vo , vertical velocity vf and angular speed w' .

By Impulse Momentum Theorem we have

Ndt = 2(vf+30)

-fdt = 2(vo-27)

Using Angular impulse angular momentum theorem about COM Of Ball

fdt*R = 2/3 *(MR^2) * (w' -10)

Here 2/3 * MR^2 Is moment of inertia of hollow spherical ball about COM

If We Assume friction to be sufficient for pure rolling on the ground we can write

vo = Rw'

Solving Above Conditions We Will Get Required Values Of vo and w'.

Also We are being given that centre of ball rises to a vertical height of 1.8 m .

Hence Using Standard equation of kinematics

vf = 6 m/sec

Now find out impulse due to friction and Normal reaction Using above equations

You Will Obtain

Ndt = 72

fdt = 17.6

We Know That Maximum Friction (On the verge of slipping) = coefficient of friction*N

So Maximum friction that ground can provide = 1/3 * 72 = 24 > 17.6

Hence The friction is sufficient to provide pure rolling on the surface.

Finally to evaluate the answer

We Use standard equation of kinematics

Time of flight = 1.2 sec( this can be obtained from maximum height information)

Range = u t + 1/2 a*t^2

Here u = 18.2 m/sec (here u is vo that we have taken above) a =9 m/sec^2 (due to electric force) t = 1.2 sec

Putting in values and solving we get

Range = 28.32 m

Well Explained Solution ,

Ayush Maurya - 5 years ago

Thanks bro! :)

Prakhar Bindal - 5 years ago

Bro! how are preparations going for Olympiads?

Prakhar Bindal - 4 years, 7 months ago

Going on ! Started with full pace from today!

Aniket Sanghi - 4 years, 7 months ago

Bro! Is Scientific Calculator allowed in NSEP and NSEA,?

Aniket Sanghi - 4 years, 6 months ago

NSEA Its surely allowed i used it last year. i dont know for NSEP

Prakhar Bindal - 4 years, 6 months ago

@Prakhar Bindal – Site shows , non-programmable calculator . The school in charge says calc with only +,-,×,÷ is allowed!

I am rather confused!

Aniket Sanghi - 4 years, 6 months ago

@Aniket Sanghi – Bro why isn't your name here?

http://vijyoshi.iiserkol.ac.in/list-of-participants.pdf

Prakhar Bindal - 4 years, 6 months ago

@Prakhar Bindal – Tried my new chem q?

Aniket Sanghi - 4 years, 6 months ago

@Aniket Sanghi – Not bro not yet

Prakhar Bindal - 4 years, 6 months ago

Electrostats + Mechanics = Fun

This is a part of my set Aniket's Mechanics Challenges

The answer is 28.32.

1 solution

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