## TomLikesPhysics 2 years ago What forces act on the block with mass m2 (as seen on the picture)? Gravity for sure but I guess the cylinder does also something. Perhaps Tension plays also some role?

1. TomLikesPhysics

2. TomLikesPhysics

The cylinder can rotate without friction and is fixed there. The moment of inertia of the cylinder is 1/2*M1*R^2.

3. cheesesticks

gravity and friction

4. cheesesticks

tension is like tug of war

5. TomLikesPhysics

Sry, I forgot to mention that there is no friction.^^

6. cheesesticks

o is it multiple choice?

7. TomLikesPhysics

Nope :D

8. Vincent-Lyon.Fr

Weight, normal force, tension in the rope.

9. cheesesticks

WAIT THATS A ROPE?

10. mahmit2012

|dw:1354312298270:dw|

11. Vincent-Lyon.Fr

That seems perfect!

12. mahmit2012

|dw:1354312672595:dw|

13. TomLikesPhysics

Do you mean with T the Tension or Torque? If you mean Tension - why is T*R=I*alpha?

14. Vincent-Lyon.Fr

T is tension. Torque is: $$\large \tau\normalsize=TR$$

15. TomLikesPhysics

So the moment of inertia of the cylinder is connected with the tension... I have never seen this formula: T=I*alpha. Where does it come from? Normally Torque = Moment of Inertia * angular acceleration but now you say it also equals a force (tension)? I think the units do not add up because normally I*alpha gives you "Nm" and not "N".

16. shamim

$m _{2}gsin \theta -T=m _{2}a$

17. shamim

$R=m _{2}gcos \theta$

18. shamim

R is a force acted by the surface on the body of mass m2

19. Vincent-Lyon.Fr

I don't think anybody wrote : T=I*alpha, where T is a tension. This is dimensionnaly wrong. Equivalent of N's 2nd law for rotating bodies is: net torque $$\Large \tau$$ = $$I\;\alpha$$

20. Vincent-Lyon.Fr

@shamim careful, R is the radius of the pulley. The problem has already been solved by mahmit2012

21. shamim

ok

22. TomLikesPhysics

Ok, but why is T*R=I*alpha? I only saw this guy when T meant Torque so I have no clue where mahmit2012 got it from.

23. Vincent-Lyon.Fr

Because torque acting on the pulley is the moment of the force $$\vec T$$ about the axis of the pulley. Moment of $$\vec T$$ is simply $$RT$$ because $$\vec T$$ acts at right angles to the radius of the pulley.

24. TomLikesPhysics

Oh ... Torque = radius cross Force. Here the force on the cylinder is the tension. Also Torque equals moment of Inertia times alpha.

25. Vincent-Lyon.Fr

Exactly!

26. TomLikesPhysics

k, I think I get it now. Thanks a lot Vincent-Lycon.Fr and mahmit2012.

27. Vincent-Lyon.Fr

yw :)) goodbye!

28. TomLikesPhysics

A shame that I can not get you booth a "Best Response". mahmit2012 solved the thing but you (Vincent-Lyon.Fr) explained it further.

29. Vincent-Lyon.Fr

NP, mahmit deserves the award.