Velocity of Pulley: Mass 1 & 2, Frictionless Disk

[Miike012]

Question: Mass one is 15.0 kg and mass 2 is 9.0 kg. They are connected by a massless rope which passes over a frictionless pulley. The pulley has a shape of a solid disk. They pulley mass is 8.0 kg and the pulley radius (R) is 0.20 meter.
What is the speed of the falling body?

Solution:

I think my issue is determining if the forces in my equations are either positive or negative...
In my diagrams I forgot to label the the direction of my force vectors...
Body 1 and Body 2: Both T1 and T2 point up towards top of page and w1 and w2 point down towards bottom of page.

Equations:

1. Ʃ (torque) = T1(R) - T2(R) = (I)(a/R) = (MR^2/2)(a/R) = M*a*R/2

2. Ʃ(F1) = T1 - w1 = (m1)(a) ... T1 = m1*a + w1

3. Ʃ(F2) = w2 - T2 = (m2)(a) ... -T2 = m2*a - w2

4. T1 - T2 = m1*a + w1 + m2*a - w2

* Substitute eq. 4. into eq. 1.

(m1*a + w1 + m2*a - w2) = M*a*/2

a = (w1 - w2)/(M/2 - m1 - m2) = -2.94 m/s^2

V = (2*-2*-2.94)^(1/2) = 3.42 m/s

The answer is 2.9 m/s

 

[ehild]

2. Ʃ(F1) = T1 - w1 = (m1)(a) ... T1 = m1*a + w1

3. Ʃ(F2) = w2 - T2 = (m2)(a) ... -T2 = m2*a - w2

Check equations 2 and 3. Object 1 accelerates downward and object 2 upward.
What was the question about speed? Do you need to give it at a certain time instant or after falling a given length?

ehild

 

[Miike012]

After it falls 2.0 meters


2. Ʃ(F1) = -T1 + w1 = (m1)(a)

3. Ʃ(F2) = -w2 + T2 = (m2)(a)

Is this correct?

 

[ehild]

Correct.

ehild