Rotation of a Ridgid Body (two masses suspended from a pulley)

[Gotejjeken]

Homework Statement

The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.55 N * m.

If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Homework Equations

Taking mass 1 (m1) to be on the left, and mass 2 (m2) to be on the right:

m1 = 4.0 kg
m2 = 2.0 kg
r = 0.06 m
Mp = 2.0 kg
f = 0.55 N * m

a(m2) = -a(m1)
T1 = m1 * a(m1) + m1 * g (T1 = Tension on mass 1)
T2 = m2 * g - m2 * a(m1) (T2 = Tension on mass 2)

Net Torque = T1 * r - T2 * r - f
alpha = Net Torque / I
I = 1/2 * Mp * r^2 (where Mp = Mass of Pulley)

a(m1) = -alpha * r (negative acceleration and positive rotation)

y = 1/2 * a * (delta)t^2

The Attempt at a Solution

Using the above equations I get:

 alpha =  2*r(T1 - T2) - f 
         ------------------
              Mp * r^2

Substituting that alpha into the a(m1) = -alpha * r equation and then solving for a(m1), I get a(m1) = -2.14 m/s^2. Then substituting this value for acceleration into the above kinematic equation and solving for time, I get the time to be 0.97s, which is wrong.

I'm not sure if my issue is just an issue of messing up the signs, or if I am approaching the problem wrong. Any help would be appreciated.

 

[alphysicist]

Hi Gotejjeken,

Homework Statement

The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.55 N * m.

If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Homework Equations

Taking mass 1 (m1) to be on the left, and mass 2 (m2) to be on the right:

m1 = 4.0 kg
m2 = 2.0 kg
r = 0.06 m
Mp = 2.0 kg
f = 0.55 N * m

a(m2) = -a(m1)
T1 = m1 * a(m1) + m1 * g (T1 = Tension on mass 1)
T2 = m2 * g - m2 * a(m1) (T2 = Tension on mass 2)

Net Torque = T1 * r - T2 * r - f
alpha = Net Torque / I
I = 1/2 * Mp * r^2 (where Mp = Mass of Pulley)

a(m1) = -alpha * r (negative acceleration and positive rotation)

y = 1/2 * a * (delta)t^2

The Attempt at a Solution

Using the above equations I get:

 alpha =  2*r(T1 - T2) - f 
         ------------------
              Mp * r^2

This might just be a typo, but here the 2 should multiply the entire numerator, not just the r(T1-T2) part.

Substituting that alpha into the a(m1) = -alpha * r equation and then solving for a(m1), I get a(m1) = -2.14 m/s^2.

It's difficult to tell (since you haven't show what numbers you used) but I believe in your calculation you are using 12cm as the radius, instead of 6cm. Is that what went wrong?

Then substituting this value for acceleration into the above kinematic equation and solving for time, I get the time to be 0.97s, which is wrong.

I'm not sure if my issue is just an issue of messing up the signs, or if I am approaching the problem wrong. Any help would be appreciated.

 

[Gotejjeken]

Oh man, not multiplying all the way through with the 2 was the problem. Thanks, I'm not sure I would have caught that and would most likely be pulling my hair out right now...sometimes it just takes an extra set of eyes =).

 

[freak_boy186]

I am doing a very similar problem right now, and I don't understand how you were able to determine what T1 or T2 were equal to.

 

[Gotejjeken]

It's been a while, but looking at my old notes I had (these are free body diagram equations):

Mass1:

(Fnet)y: T1 - m1*g = m1*a(m1)y

Mass2:

(Fnet)y: T2- m2*g = m2*a(m2)y

Then just solved those for T1 and T2 respectively. After that it looks like an application of the torque equations and then putting everything together to get that final equation above and getting acceleration / falling time from there.