- #1
rainstom07
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Homework Statement
In the figure http://i.imgur.com/Y5Vc7.gif shows a uniform disk (M = 2.5kg, R = 0.20 m) mounted on a fixed horizontal axle. A block (m = 1.2kg) hangs from a massless cord that is wrapped around the rim of the disk. The cord does not slip and there is no friction.
Find the acceleration of the block.
Homework Equations
[tex]\tau_{net} = I\alpha[/tex]
[tex]a_t = \alpha r[/tex]
The Attempt at a Solution
1. [tex]F_{net} = -ma = T - mg[/tex]
2. [tex]a = -\frac{T}{m} + g[/tex]
3. [tex]\tau_{net} = I\alpha = \frac{1}{2}MR^2\alpha = -TR[/tex]
4. [tex]\alpha = \frac{a_t}{r} = \frac{a}{r}[/tex] the tangential acceleration is also the linear acceleration.
5. [tex]-TR = \frac{1}{2}MR^2(\frac{a}{r})[/tex]
6. [tex]T = -\frac{1}{2}Ma[/tex]
Plugging equation 6 into 2 yields:
[tex]a = \frac{Ma}{2m}+g[/tex]
Some manipulation
[tex]a = \frac{g}{1-\frac{M}{2m}}[/tex]
[tex]a = \frac{9.8}{1-\frac{2.5}{2*1.2}} = -235.2[/tex]
This is obviously wrong.
5. Solution
7. [tex]F_{net} = ma = T - mg[/tex]
8. [tex]a = \frac{T}{m} - g[/tex]
plugging equation 6 into equation 8 yields:
[tex]a = -\frac{g}{1+\frac{M}{2m}}[/tex]
[tex]a = -\frac{g}{1+\frac{2.5}{2*1.2}} = -4.8 ms[/tex]
6. My question
Obviously my model for the net force is wrong. I used -ma instead of ma.
Why is it okay for me to consider the direction of the net torque to be -TR instead of TR for equation 3, but it isn't okay for me to consider the direction of the net force to be -ma instead of ma for equation 1? The net force is pointing downward isn't it?
Also vice-versa
Why is it okay for me to consider the direction of the net force to be ma instead of -ma for equation 1, but it isn't okay to consider the direction of the net torque TR instead of -TR for equation 3? shouldn't i be consistent?
Thanks in advance. This stuff always gets me in physics.
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