# Atwood Machine Problem (Rotational Dynamics)

• Feb 19th 2010, 11:35 AM
apple123
Atwood Machine Problem (Rotational Dynamics)
Two masses m1 and m2 are hung over a pulley, forming an Atwood Machine.

m1 = 0.400 kg
m2 = 0.410 kg
R, radius of pulley = 0.10 m
moment of inertia I of pulley = 5E-3 kgm^2.

a) When the distance between the two masses is 0.50 m, how fast is each moving?
b) At that time, how fast would each mass move if the pulley was considered massless?

Well for part a I thought that:
(m1)a = T - (m1)g
(m2)a = (m2)g - T

and just solve for a. Then from a you can find the speed when the mass has moved 0.25 m.

However I would have done the exact same thing for part b, and I didn't use moment of inertia here at all. So I'm sure this is wrong... any help?
• Feb 19th 2010, 02:58 PM
Roam
Quote:

Originally Posted by apple123
Two masses m1 and m2 are hung over a pulley, forming an Atwood Machine.

m1 = 0.400 kg
m2 = 0.410 kg
R, radius of pulley = 0.10 m
moment of inertia I of pulley = 5E-3 kgm^2.

a) When the distance between the two masses is 0.50 m, how fast is each moving?
b) At that time, how fast would each mass move if the pulley was considered massless?

Well for part a I thought that:
(m1)a = T - (m1)g
(m2)a = (m2)g - T

and just solve for a. Then from a you can find the speed when the mass has moved 0.25 m.

However I would have done the exact same thing for part b, and I didn't use moment of inertia here at all. So I'm sure this is wrong... any help?

Are we to assume the pulley is both massles and frictionless? Mass 2 is heavier than mass 1. It is convinient in situations involving objects connected by cords to take the direction of motion as positive direction. In this case, for $\displaystyle m_1$ we take up positive, and down positive for $\displaystyle m_2$. (If we do this the acceleration will be positive for each mass).

Writing $\displaystyle \sum F_y = ma_y$ for each mass in turn we have

$\displaystyle F_T - (0.4)(9.81)N = (0.4)a$

And for the second mass, $\displaystyle (0.41)(9.81)N-F_T = (0.41)a$

If we add the two equations, the unknown $\displaystyle F_T$ drops out giving

$\displaystyle (0.41-0.4) (9.81) =0.81 a$

For which $\displaystyle a=0.12 m/s^2$

Maybe this is how the problem needs to be approached.
• Feb 20th 2010, 09:54 AM
skeeter
Quote:

Originally Posted by apple123
Two masses m1 and m2 are hung over a pulley, forming an Atwood Machine.

m1 = 0.400 kg
m2 = 0.410 kg
R, radius of pulley = 0.10 m
moment of inertia I of pulley = 5E-3 kgm^2.

a) When the distance between the two masses is 0.50 m, how fast is each moving?
b) At that time, how fast would each mass move if the pulley was considered massless?

Well for part a I thought that:
(m1)a = T - (m1)g
(m2)a = (m2)g - T

and just solve for a. Then from a you can find the speed when the mass has moved 0.25 m.

However I would have done the exact same thing for part b, and I didn't use moment of inertia here at all. So I'm sure this is wrong... any help?

Let $\displaystyle T_1$ = tension in the cord on the $\displaystyle m_1$ side of the pulley

$\displaystyle T_2$ = tension in the cord on the $\displaystyle m_2$ side of the pulley

scalar equations for the net force on each side ...

$\displaystyle m_2g - T_2 = m_2a$

$\displaystyle T_1 - m_1g = m_1a$

Newton's 2nd law for rotation ...

$\displaystyle T_2 R - T_1 R = I \alpha = I \frac{a}{R}$

solve the system for "a"