2 masses, a massive pulley and an inclined surface

1. Sep 7, 2016

Karol

1. The problem statement, all variables and given/known data
A mass m lies on an inclined surface with equal coefficients of friction μ=μsk. the pulley has also mass m and the weight is 2m. what is the velocity after it has descended distance h and during how much time.

2. Relevant equations
Moment of inertia of a massive disk: $I_{cen}=\frac{1}{2}mr^2$
Torque and angular acceleration: $M=I\alpha$

3. The attempt at a solution
$$\Sigma F=ma:~~\left\{\begin{array}{l} 2mg-T_1=2ma \\ T_2-mg\sin\theta-mg\mu\cos\theta=ma \end{array}\right.$$
$$\rightarrow~T_1=\frac{2}{3}(1+\sin\theta+\mu\cos\theta)mg$$
$$T_2=\frac{mg}{9}(4+7\sin\theta+7\mu\cos\theta)$$
$$M=I\alpha:~~a=\frac{2}{m}(T_1-T_2)=...=\frac{2g}{3}\left( \frac{2}{3}-\frac{1}{3}\sin\theta-\frac{\mu}{3}\cos\theta \right)$$

2. Sep 7, 2016

Simon Bridge

Well done - I have not checked the algebra but that looks like a good approach to take.
Is there a question in all that?

3. Sep 7, 2016

Karol

Thank you Simon, i am not sure about my answer, that is why i posted. thank you

4. Sep 8, 2016

ehild

I got different result. T1 and T2 are not needed. Eliminate them by adding the force equations and substituting T1-T2 from the torque equation.

5. Sep 8, 2016

haruspex

Looks like you divided by 3 twice over.

6. Sep 8, 2016

Karol

$$\left\{\begin{array}{l} (\rm 1)~2mg-T_1=2ma \\ (\rm 2)~T_2-mg\sin\theta-mg\mu\cos\theta=ma \end{array}\right.$$
$$(\rm 1)+(\rm 2),~T_2-T_1=-\frac{1}{2}ma:~~a=\frac{2}{7}(2-\sin\theta-\mu\cos\theta)g$$

7. Sep 8, 2016

Correct!

8. Sep 8, 2016

Karol

Thanks Ehild, Simon and Haruspex

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted