Bouncing Balls

Classical Mechanics Level 5

A solid spherical ball of radius 5m is projected to the right from a rough ground with a velocity 10 m/s at an angle $\theta$ with the horizontal. Also at the same time it is given an angular velocity of 3 rad/s clockwise such that axis of rotation is perpendicular to plane of projectile .

Let horizontal distance travelled by the ball from the point of projection to the point where it made it's second bounce be $D$ . Given that maximum value of $D$ occurs at $\theta = \arccos { \frac { (\sqrt { a } -b) }{ c } }$ where $a,b,c$ are integers, $a$ is square free, $b,c$ are co-prime then find $a+b+c$ .

Assumptions and details

Co-efficient of restitution of ground = 0.5
Co-efficient of friction of ground with the ball is $\frac { 1 }{ 3 }$
Mass of the ball = 5 kg
Take $g=10 \text{ m/s}^2$

The answer is 2976.

1 solution

Ronak Agarwal
Jun 13, 2014

Distance travelled by particle before first bounce ${ R }_{ 1 }=\frac { { v }^{ 2 } }{ g } sin(2\theta )=10sin(2\theta )$

As the ball bounces for the first time it is easy to note that the point of contact of the ball is tending to go backward hence friction will be forward direction.

${ v }_{ y }(initial)=-10\sin { \theta } j$

Since coefficient of restitution is $0.5$ hence ${ v }_{ y }(final)=5\sin { \theta } j$

We say let impulse due to friction be $f$

If we assume friction to be sufficient then :

${ \omega }_{ final }R={ v }_{ x }(final)$ ) (i)

Writing momentum-impulse equations:

$f=m({ v }_{ x }(final)-10cos\theta )$ (ii)

$fR=\frac { 2 }{ 5 } m{ R }^{ 2 }(3-{ \omega }_{ final })$ (iii)

Solving ${ v }_{ x }(final)= (50\cos { \theta } +6R)/7$

Since $R=5, { v }_{ x }(final)=(50\cos { \theta } +30)/7$

Finally distance between point of first bounce to second bounce is :

${ R }_{ 2 }=\frac { 2{ v }_{ y }(final){ v }_{ x }(final) }{ g } =\frac { 10sin\theta }{ g } (\frac { 50cos\theta +30 }{ 7 } ) =(25sin2\theta +30sin\theta )/7$

Hence $D={ R }_{ 1 }+{ R }_{ 2 }=10sin2\theta +\frac{25sin2\theta +30sin\theta }{7}=\frac{95sin2\theta +30sin\theta}{7}$ .

Now to maximise this we differentiate D w.r.t $\theta$ and equate it to $0$

$\Longrightarrow 190cos2\theta +30cos\theta = 0$

$\Rightarrow 38{ cos }^{ 2 }\theta +3cos\theta -19=0$

$Solving cos\theta = \frac { (\sqrt { 2897 } -3) }{ 76 }$

Hence $\boxed{a=2897,b=3,c=76}$

We get $a+b+c=2976$

Sorry for this very long solution guys!

Please comment if you believe the solution to be wrong and please upvote it if you believe it to be right

Ronak Agarwal - 6 years, 12 months ago

You should have informed if ball is going right or left, because the ball's rotation can be help or disturb the ball's movement.

But thanks for more this question, I'll keep it in my favorite set!

Felipe Hofmann - 6 years, 10 months ago

I believed that the convention is assumed to be right. Anyway I will put up a diagram to the question

Ronak Agarwal - 6 years, 10 months ago

Same but I use angular momentum conservation about bottom most point.

$(10\cos { \theta } )r\quad +\quad \cfrac { 2 }{ 5 } m{ r }^{ 2 }{ \omega }_{ o }\quad =\quad { V }_{ x }^{ " }r\quad +\quad \cfrac { 2 }{ 5 } m{ r }^{ 2 }{ \omega }^{ " }\quad$ .

Nice question @Ronak agarwal ! But the most important part of this question is that to think that sphere Can Rolls purely .

And I think You should state in question that there is sufficient friction instead of giving coefficient of friction to remove ambiguity in question

Deepanshu Gupta - 6 years, 7 months ago

You are missing one thing @Deepanshu Gupta here, first thing there is no ambiguity in the question, in fact you have to check whether the case of maximum range there is sufficient friction or not, I have checked it while making this question and it came out to be sufficient. Don't underappriciate this part of the question.

Ronak Agarwal - 6 years, 7 months ago

Oh Okay I got it ! Amazing !!
. Really This is beautiful part of this question! Thanks for notifying me for this. @Ronak Agarwal

Deepanshu Gupta - 6 years, 7 months ago

Could you please help me out? This is how I proceeded. After finding out that the angular velocity will make the friction point in the right direction during the first bounce, we also see that the angular momentum decreases.

$Let\quad N\quad be\quad the\quad vertical\quad force\quad that\quad acts\quad on\quad the\quad ball\quad on\quad the\quad first\quad bounce.\\ N=\frac { dP }{ dt } =\frac { \Delta P }{ \Delta t } \\ We\quad know,\quad { V }_{ fv }=\frac { { V }_{ iv } }{ 2 } \quad \quad \quad (v\quad for\quad vertical)\\ \Delta P=\frac { 3mv\sin { \theta } }{ 2 } \quad (taking\quad signs\quad into\quad consideration)\\ N\Delta t=\frac { 3mv\sin { \theta } }{ 2 } \\ Now,\quad frictional\quad force\quad =\frac { N }{ 3 } \\ \tau =\frac { Nr }{ 3 } =\frac { \Delta L }{ \Delta t } =\frac { I({ w }_{ f }-{ w }_{ i }) }{ \Delta t } or\frac { I({ w }_{ i }-{ w }_{ f }) }{ \Delta t } \quad \quad (Since\quad { w }_{ i }>{ w }_{ f })\\ N\Delta t=3\times \frac { 2mr }{ 5 } \times (3-{ w }_{ f })\\ We\quad get,\\ \frac { 3mv\sin { \theta } }{ 2 } =\frac { 6mr(3-{ w }_{ f }) }{ 5 } \\ Which\quad gives\quad us,\quad \\ { w }_{ f }=3-\frac { 5\sin { \theta } }{ 2 } \\ Then\quad we\quad use\quad conservation\quad of\quad angular\quad momentum\quad about\quad the\quad \\ point\quad of\quad contact\quad to\quad find\quad horizontal\quad component\quad of\quad { V }_{ f }$

I didn't understand how one can assume the sphere would go into rolling motion right after its first bounce.

Dhruva Patil - 5 years, 6 months ago

Those who have the doubt how I have assumed friction to be sufficient,you can derive the expression for fricional impulse and check it is less than the maximum possible friction.

Ronak Agarwal - 6 years, 11 months ago

can u do it plz

Dhruv Aggarwal - 5 years, 1 month ago

wat if slipping took place during contact with ground??\

Mukesh Raghav - 6 years, 10 months ago

That is what I wanted to say.If the frictional impulse required
for no slipling is more than the maximum frictional impulse,than sliping would take place with the ground.

Ronak Agarwal - 6 years, 10 months ago

@Ronak Agarwal – But it can also be that the friction is not sufficient to start pure rolling.?

Tushar Gopalka - 6 years, 7 months ago

Ronak u have a great problem solving capability, keep it up. What is your daily routine ,how long u study.

Anmol Mahapatra - 6 years, 10 months ago

I'm not very good at rotational motion - how does one arrive at the momentum-impulse equations? Thanks!

Shaun Leong - 5 years, 3 months ago

One should approach those when they see that collision is taking place

Md Zuhair - 3 years, 5 months ago

Silly me... got correct at the 3rd attempt

Md Zuhair - 3 years, 5 months ago

Hi Bro. See my profile. There 3rd or 4th question from the top is your related to collision and spring. There the friction required for pure rolling is coming out to be less than the maximum. But I am getting answer when I take the maximum value of frictional impulse. Please check.

Kushagra Sahni - 3 years, 5 months ago

Bouncing Balls

The answer is 2976.

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