A solid, frictionless cylindrical pulley has a radius R = 0.550 m. A block is at

Question

A solid, frictionless cylindrical pulley has a radius R = 0.550 m. A block is attached to a cord that is wrapped around the outside of the pulley. The block is falling with a downward acceleration of 4.20 m/s2, and the tension in the cord is 0.800 N. The free body diagrams for the pulley and the block are shown below. Note that the magnitudes of the two tensions T1 and T2 are equal. (Just call the tension T.) Also, "n" in the free body diagram of the pulley represents the normal force provided by the axle of the pulley, equal in magnitude to the weight of the pulley. Note that the normal force and the weight force do not produce torques on the pulley. (Why is this?)

(a) Determine the angular acceleration of the pulley.
rad/s2

(b) Determine the moment of inertia of the pulley. HINT: You will need to use the expressions for torque.
kg m2

(d) Determine the mass of the block.
kg

Explanation / Answer

Given data

radius=0.550m

accelaration of the block=4.2m/s2

tension=0.8N

A)Angular accelaration of the pulley

a=r*alfa

alfa=a/r=4.2/0.55=7.63rad/s2

B)moment of inertia of the pulley

torque=rXForce

torque=I*alfa

equate both torques

r*Force=I*alfa

I=r*force/alfa

I=0.550*0.8/7.63

I=0.05kg-m2

C)from the free body diagrams

we can write as

ma=mg-T

T=m(g-a)

m=T/(g-a)=0.8/5.8=0.137kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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