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Pulley problem (Rotation)

  • Thread starter greyradio
  • Start date
12
0
[SOLVED] Pulley problem (Rotation)

Mass m1 = 9.07 kg sits on a frictionless, horizontal surface. A massless string is tied to m1, passes over a pulley (a solid disk of mass Mp = 5.42 kg and radius Rp = 37.9 cm), and is tied to m2 = 6.94 kg hanging in space.

Find the acceleration of the masses.

It seems I'm having some trouble with this problem. I'm not sure if my algebra is right or if I am doing the wrong thing with the moment of Inertia. I was hoping someone could help me.

torque = I [tex]\alpha[/tex]
T = ma
a = R[tex]\alpha[/tex]


My attempt is as follows:

The question assumes that there are non-slip conditions and it starts from rest.

1 . First, I find the pulley's torque

T1 = ma
T2 = mg - ma
torque = I [tex]\alpha[/tex]
r T = I [tex]\alpha[/tex]
T2 R - T1 R = I [tex]\alpha[/tex]

2. Second, I find the acceleration.

a = R [tex]\alpha[/tex]
a / R = [tex]\alpha[/tex]

3. Third, I solve for a.

T2 R - T1 R = I[tex]\alpha[/tex]
R (T2 - T1) = I (a/R)
T2 - T1 = I (a/R) / R
T2 - T1 = I a / R[tex]^{2}[/tex]
m2g-m2a - m1a = I a/R[tex]^{2}[/tex]
m2g = m2a + m1a + I a/R[tex]^{2}[/tex]
m2g = a (m2 + m1 + I /R[tex]^{2}[/tex]
m2g / m2 + m1 + I / R[tex]^{2}[/tex] = a
(6.49)(9.81) / (I/R[tex]^{2}[/tex]) + 9.07 +6.94 = a
I = mass of pulley * radius[tex]^{2}[/tex]
I = (5.42 kg)(.379 m)[tex]^{2}[/tex]
I = .7785

(6.49)(9.81) / (.7785/.379[tex]^{2}[/tex]) + 9.07+6.94 = a
63.667 / 5.420 + 9.07 + 6.94 =
63.667/21.43 = 2.971 m/s[tex]^{2}[/tex]
 
Last edited:

Answers and Replies

454
0
It's OK except for the moment of inertia of a disk. it's I = (1/2)mR^2.
I would also use more parentheses and write m2g / (m2 + m1 + I / R^2) instead of
m2g / m2 + m1 + I / R^2
 
12
0
yeah i'm sorry i had to edit it a few times since some of the subscripts were displayed oddly. It worked out it thanks a lot for your help.
 

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