Asked by cindy
A weather vane initially at rest has a moment of inertia of 0.115 kg · m2 about its axis of rotation. A 44.5 g piece of clay is thrown at the vane and sticks to it at a point 10.5 cm from the axis. The initial velocity of the clay is 18.5 m/s, directed perpendicular to the vane. Find the angular velocity of the weather vane just after it is struck.
Answers
Answered by
Elena
The moment of inertia of weather vane with piece of clay is
I = I(vane) +mR^2 = 0.115+44.5•10^-3•(0.105)^2=
= 1.2•10^-2 kg • m2.
The torque due to the clay hitting is M=F•R
F•t=Δp=m•v, therefore F=m•v/t.
Then M=F•R=m•v•R/t.
From Newton’s 2 Law for rotation M=I• ε.
m•v•R/t =I•ε.
m•v•R/I= ε• t = ω
ω = m•v•R/I ==44.5•10^-3•18.5•0.105/1.2•10^-2= 7.21 rad•s
I = I(vane) +mR^2 = 0.115+44.5•10^-3•(0.105)^2=
= 1.2•10^-2 kg • m2.
The torque due to the clay hitting is M=F•R
F•t=Δp=m•v, therefore F=m•v/t.
Then M=F•R=m•v•R/t.
From Newton’s 2 Law for rotation M=I• ε.
m•v•R/t =I•ε.
m•v•R/I= ε• t = ω
ω = m•v•R/I ==44.5•10^-3•18.5•0.105/1.2•10^-2= 7.21 rad•s
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