A string passing over a pulley has a 4.00-kg mass hanging from one end and a 2.6

Question

A string passing over a pulley has a 4.00-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.1cm and mass 0.72kg .

If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?

Part B

In fact, it is found that if the heavier mass is given a downward speed of 0.30m/s , it comes to rest in 6.6s . What is the average frictional torque acting on the pulley?

Express your answer using two significant figures.

Explanation / Answer

Let the tension on the heavier mass's side be T1 and on the lighter mass's side be T2.

4*9.81-T1=4a

T2-2.6*9.81=2.6a

And for rotational forces on the pulley,

(T1-T2)*0.041=I*alpha=0.5*0.72*0.041^2*a/0.041

(T1-T2)=0.36a

Solving simultaneously,

a=1.97m/s^2

b)a=0.3/6.6=0.0454

4*9.81-T1=0.0454*4

T2-2.6*9.81=2.6*0.0454

Let b be the frictional torque.

(T1-T2)*0.041-b=0.36*0.0454*0.041

Solving for b,

b=0.55Nm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.