# Friction acting on the shot putter

Tags:
1. Oct 31, 2016

### decentfellow

1. The problem statement, all variables and given/known data
A shot putter with a mass of $80$ kg pushes the iron ball of mass of $6$ kg from a standing position accelerating it uniformly from rest at an angle of $45^{\circ}$ with the horizontal during an interval of $0.1$ seconds. The ball leaves his hand when it is $2\text{m}$ high above the level ground and hits the ground $2$ seconds later.

The minimum value of coefficient of friction if the shot putter does not slip during the shot is closest to:-
a) $0.28$
b) $0.38$
c) $0.48$
d) $0.58$

2. Relevant equations

$\Delta S= u_yt-\dfrac{gt^2}{2} \tag{1}$
$f_{max}=\mu_{min}N\tag{2}$

3. The attempt at a solution
First of all we need to find the acceleration of the ball. For that we use the fact that we are given with the time for which a force was applied in the ball which is $0.1$ $\text{sec}$, the time of flight of the ball and the height of projection of the ball.

So, we have
• Time for which a force had been applied on the ball $=0.1$ $\text{seconds}$
• Height of projection of the iron ball $=2$ $\text{m}$
• Angle of projection $=45^{\circ}$
• Time of flight of the iron ball $=2$ $\text{seconds}$
So, initial velocity of the ball $=u=at=0.1\times a$

From $\text{eq}^{\text{n}}$ $(1)$, we get
$$\Delta S= u_yt-\dfrac{gt^2}{2}\implies -2=\left[(a\times 0.1)\times \sin{45^{\circ}}\times 2\right]-(5\times 4)\implies a=90\sqrt2\text{m/s}^2$$

So, the force exerted by the iron ball on the shotputter is $F=6\times 90\sqrt2=540\sqrt2$. Now consider the FBD of the shotputter as shown below:-

From the FBD we get,

$N=F\sin\theta + Mg$
$f_s=F\cos\theta\implies \mu_{min}N=F\cos\theta\implies \mu_{min}=\dfrac{F\cos\theta}{Mg+F\sin\theta}=\dfrac{540}{800+540}=\dfrac{27}{67}\approx 0.40$

Though the question asks for the nearest value and 0.40 is nearest to 0.38, so it should be the answer and so is the case. But I am not sure if I am doing everything correct or not because my answer has a difference of 0.02 which is a considerable difference considering the options that are given.

2. Oct 31, 2016

### TomHart

I worked out the problem without considering your solution and got μ = 0.402.
I used g = 9.8 ms-2

Standard qualifier: I make lots of mistakes so don't bet the farm on my answer.

3. Oct 31, 2016

### decentfellow

So, my solution is also correct and I should consider the answer $0.38$ to be the correct one.

4. Nov 1, 2016

### haruspex

Did gravity suddenly start acting on the shot at the moment it was released?

5. Nov 1, 2016

### decentfellow

I think you are telling me to incorporate the action of gravitational force in the motion of the ball before the shotputter releases the iron ball. Is that it? If so I think that the question has already incorporated it and that's why it tells us that the shotputter exerts force and doesn't expect of us to consider the action of the gravitational force on the ball. I know that I am not at all clear but I think this is the best that I can do.

6. Nov 1, 2016

### haruspex

You are not considering how that affects the normal force between the athlete and the ground.

7. Nov 1, 2016

### decentfellow

Oh thank you for pointing that out...you saved my life. Had been banging my head for quite some time. Yes, you are right that was precisely what I was missing and the answer comes out to be $0.385$. So that settles it.