1. ## ball rolling

I’ve come across a physics problem and attempted to solve it, but I’ve forgotten a lot of it and haven’t arrived at a solution. See the attached pic.

Can someone guide me in the right direction or solve it?

I’ve tried

sin 30º = h / PQ
1/2 = 2 m / PQ
PQ = 4 m

Fg * cos 30º * 4m * μ = Fg * x * μ
cos 30º * 4m = x
sqrt(3)/2 * 4m = x
x = 2 * sqrt(3)

Anyhow, I fail to obtain μ and I feel like my approach is wrong and x is thus also wrong.

2. ## Re: ball rolling

Actually you are doing OK so far. The energy change of the system is equal to the decrease in PE as the ball descends from height h:

$\Delta PE = mg h$

The work done by friction comes in two parts. First is friction force on the incline of $\mu mg \cos \theta$ operating over the distance of the incline, which is $\frac h {\sin \theta}$. The second component of work is the friction force along the horizontal section times the distance it slides until it stops, or $\mu mg x$. So the full equation is:

$mgh = \mu mg \cos \theta \frac h {\sin \theta} + \mu mg x$

You already have the value for x, so substitute that value into this equation and solve for $\mu$

3. ## Re: ball rolling

Thanks, it makes sense now.

I’m just a little puzzled about x.
If I try μ = 0.2 and x = 6.5 m (answer (a)), your equation works.
It looks like I did something wrong when I tried to work out the horizontal section (x).

4. ## Re: ball rolling

Sure those values satisfy one of the equations, but do they satisfy the fact that work done by friction on the incline = the work done by friction on the horizontal section? There is nothing wrong with your answer for 'x'.

5. ## Re: ball rolling

If I use my ’x’ then the result for μ is:

h = 2
x = 2 * sqrt(3) ≈ 3.464
--
mgh = μgh cosΦ * (h/sinΦ) + μmgx
2 = 2μ * (sqrt(3)/2) * 4 + μ * 2 * sqrt(3)
μ = sqrt(3)/9 = 0.192

I’m guessing x = 3.464 m and μ = 0.192 are good enough for me to choose answer (2) then - 3.5m and 0.2 friction coefficient.

6. ## Re: ball rolling

Originally Posted by metlx
If I use my ’x’ then the result for μ is:

h = 2
x = 2 * sqrt(3) ≈ 3.464
--
mgh = μgh cosΦ * (h/sinΦ) + μmgx
You have included 'h' twice in the first part of the right hand side, instead of 'm'. This should be:

$mgh = \mu mg \cos \theta \left( \frac {h}{\sin \theta} \right) + \mu mgx$

Try it again from here.

7. ## Re: ball rolling

Ooops. I hope it’s right this time.

$\mu = \frac{\sqrt{3}}{6} \approx 0.29$

(3) then.

8. ## Re: ball rolling

Originally Posted by metlx
$\mu = \frac{\sqrt{3}}{6} \approx 0.29$

(3) then.
Bingo!