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Two thin rectangular sheets (0.06 m 0.52 m) are identical. In the first sheet the axis of rotation lies along the 0.06 m side,...Question
Two thin rectangular sheets (0.09 m 0.46 m) are identical. In the first sheet the axis of rotation lies along the 0.09-m side, and in the second it lies along the 0.46-m side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in 7.5 s. How long does it take for the second sheet, starting from rest, to reach the same angular velocity?
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Elena
The moment of inertia of the rectangular plate about its side is
I =ma²/3 , where a is the length of another side (perpendicular to the axis of rotation), therefore,
I1 =m•(0.46)²/3,
I2 =m•(0.09)²/3,
Final angular velocity is
ω = ε1•t1, ε1 =ω /t1
ω = ε2•t2, • ε2 =ω /t2
The Newton’s 2 Law for rotation
M=I•ε,
M=I1• ε1 = I1• ω /t1= m•(0.46)² • ω /3•t1,
M=I2• ε2= I2• ω /t2 = m•(0.09)² • ω /3•t2,
m•(0.46)² • ω /3•t1 = m•(0.09)² • ω /3•t2,
(0.46)² /t1 =(0.09)² /t2,
t2 =t1• (0.09/0.46)² = 7.5•3.8•10^-2= 0.28 s.
I =ma²/3 , where a is the length of another side (perpendicular to the axis of rotation), therefore,
I1 =m•(0.46)²/3,
I2 =m•(0.09)²/3,
Final angular velocity is
ω = ε1•t1, ε1 =ω /t1
ω = ε2•t2, • ε2 =ω /t2
The Newton’s 2 Law for rotation
M=I•ε,
M=I1• ε1 = I1• ω /t1= m•(0.46)² • ω /3•t1,
M=I2• ε2= I2• ω /t2 = m•(0.09)² • ω /3•t2,
m•(0.46)² • ω /3•t1 = m•(0.09)² • ω /3•t2,
(0.46)² /t1 =(0.09)² /t2,
t2 =t1• (0.09/0.46)² = 7.5•3.8•10^-2= 0.28 s.