Asked by Jessica
There are two spinning disks A and B arranged so that they initially are separated but can be pushed into contact. Disk A is made of a lighter material, so it has a moment of inertia that is one third that of disk B. Disk A starts off spinning at some ùi when the two disks are separated. The two disks are then pushed together, coupling their motion. 2400 J of heat energy were produced in this process. What was the initial kinetic energy of disk A?
Answers
Answered by
Elena
J1=3•J , initial angular velocity =ω, the angular velocity of two discs=ω1
The law of conservation of angular momentum
L1=L2
J •ω= J •ω1 +J 1• ω1=
= ω1(J+3•J)=4•J •ω1.
=> ω1= ω/4.
KE1 =J •ω²/2,
KE2= J• ω1²/2+ J1• ω1²/2 =
=J• ω1²/2+ 3•J• ω1²/2 =4•J •ω1²/2.
ΔKE=KE2-KE1=4•J •ω1²/2 - J •ω²/2=
=(J/2) •(4 •ω1²- ω²)=
=(J/2) •(4 •ω²/16 - ω²)=
=3J•ω²/8 =(3/4)( J•ω²/2) =2400 J
KE1J•ω²/2 = 4•2400/3= 3200 J.
The law of conservation of angular momentum
L1=L2
J •ω= J •ω1 +J 1• ω1=
= ω1(J+3•J)=4•J •ω1.
=> ω1= ω/4.
KE1 =J •ω²/2,
KE2= J• ω1²/2+ J1• ω1²/2 =
=J• ω1²/2+ 3•J• ω1²/2 =4•J •ω1²/2.
ΔKE=KE2-KE1=4•J •ω1²/2 - J •ω²/2=
=(J/2) •(4 •ω1²- ω²)=
=(J/2) •(4 •ω²/16 - ω²)=
=3J•ω²/8 =(3/4)( J•ω²/2) =2400 J
KE1J•ω²/2 = 4•2400/3= 3200 J.
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