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

 

[Nickie]

To find the acceleration of the two masses in this pulley system, you can use the equations of rotational motion and Newton's second law. As you have correctly identified, the torque equation is Ʃτ=Iα, where Ʃτ is the net torque, I is the moment of inertia of the pulley, and α is the angular acceleration. In this case, the net torque would be equal to (T1-T2)r, where r is the radius of the pulley.

Next, you can use the equations T1-mAg=mAa and mBg-T2=mBa, where T1 and T2 are the tensions in the string on either side of the pulley, m is the mass of each hanging object, a is the acceleration, and g is gravity. These equations represent the forces acting on each mass - the tension of the string pulling them up, and their weight pulling them down.

By substituting the expression for net torque into the torque equation, you can solve for the angular acceleration α. Then, using this value of α, you can solve for the acceleration a using either of the two equations involving the tensions and masses.

It's also worth noting that in this system, both masses will have the same acceleration since they are connected by the same string. So you can solve for the acceleration of one mass and it will be the same for the other.

I hope this helps. Good luck with your calculations!