Rotational motion on pulley system

[NathanLeduc1]

Homework Statement

A string passing over a pulley has a 3.80-kg mass hanging from one end and a 3.15-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.0 cm and mass 0.80 kg. If the bearings of they pulley were frictionless, what would be the acceleration of the two masses?

Homework Equations

I=0.5mr²
a=rα
Ʃτ=Iα
τ=Fr
T₁-m_Ag=m_Aa
m_Bg-T₂=m_Ba

The Attempt at a Solution

I tried rearranging the bottom two equations into the form:
a=(T₁-m_Ag)/m_A
a=(m_Bg-T₂)/m_B

I then plugged variables into the following equation:
Ʃτ=0.5mr₂α
(T₂-T₁)r=0.5mr₂α
(T₂-T₁)r=0.5mr₂(a/r)

This equation then simplifies to:
a=(2T₂-2T₁)/m

This is where I'm stuck. How do I proceed from here? Thanks.

 

[ehild]

Homework Statement

A string passing over a pulley has a 3.80-kg mass hanging from one end and a 3.15-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.0 cm and mass 0.80 kg. If the bearings of they pulley were frictionless, what would be the acceleration of the two masses?

Homework Equations

I=0.5mr²
a=rα
Ʃτ=Iα
τ=Fr
T₁-m_Ag=m_Aa
m_Bg-T₂=m_Ba

The Attempt at a Solution

I tried rearranging the bottom two equations into the form:
a=(T₁-m_Ag)/m_A 1
a=(m_Bg-T₂)/m_B 2

I then plugged variables into the following equation:
Ʃτ=0.5mr₂α
(T₂-T₁)r=0.5mr₂α
(T₂-T₁)r=0.5mr₂(a/r)

This equation then simplifies to:
a=(2T₂-2T₁)/m

This is where I'm stuck. How do I proceed from here? Thanks.

Isolate T1 from 1 and T2 from 2 and substitute them into the last equation.

ehild

 

[NathanLeduc1]

Okay, I can do that but then how do I solve for T1 and T2?

 

[ehild]

Okay, I can do that but then how do I solve for T1 and T2?

Substitute the numerical value for a in equations 1) and 2).

ehild