Rotational motion on pulley system

  • #1

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-mAg=mAa
mBg-T2=mBa

The Attempt at a Solution


I tried rearranging the bottom two equations into the form:
a=(T1-mAg)/mA
a=(mBg-T2)/mB

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.
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,914

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-mAg=mAa
mBg-T2=mBa

The Attempt at a Solution


I tried rearranging the bottom two equations into the form:
a=(T1-mAg)/mA 1
a=(mBg-T2)/mB 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
 
  • #3
Okay, I can do that but then how do I solve for T1 and T2?
 
  • #4
ehild
Homework Helper
15,543
1,914
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
 

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