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Todd asked in Science & MathematicsPhysics · 7 days ago

PHYSICS ASAP!?

A playground merry-go-round of radius 1.8 m

has a moment of inertia 239 kg.m2

and is

rotating at 15 rev/min about a frictionless

vertical axle. Facing the axle, a 21 kg child

hops onto the merry-go-round, and manages

to sit down on the edge.

What is the new angular speed of the

merry-go-round?

Answer in units of rev/min.

BONUS: A cable passes over a pulley. Because the

cable grips the pulley and the pulley has nonzero mass, the tension in the cable is not the

same on opposite sides of the pulley. The

force on one side is 138 N, and the force on

the other side is 79 N.

Assuming that the pulley is a uniform disk

of mass 2.68 kg and radius 1.113 m, find the

magnitude of its angular acceleration. [For a

uniform disk, I = (1/2)mr2

.]

Answer in units of rad/s

2

.

1 Answer

Relevance
  • oubaas
    Lv 7
    7 days ago

    1)

    before

    Jmgr = 239 kg*m^2

    Jmgr  = 2PI/60*n = 0.10472*15 = 1.57 rad/sec 

    L = J*ω = 239*1.57 = 375 kg*m^2/sec

    afterJc = mc*r^2 = 21*1.8^2 = 68 kg*m^2

    J = Jmgr+Jc = 239+68 = 307 kg*m^2

    angular momentum L is conserved 

    ω' = L/J = 375/307 = 1.22 rad/sec 

    2)

    MoI J = m/2*r^2 = 1.34*1.113^2 = 1.660 kg*m^2

    ΔT*r = J*α

    angular acceleration α = 59*1.113/1.660 = 39.56 rad/sec^2

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