What torque would have to be applied to a rod with a length of 5 m5m and a mass of 2 kg2kg to change its horizontal spin by a frequency of 9 Hz9Hz over 6 s6s?

1 Answer
Feb 21, 2016

L=(25*pi)/2L=25π2

Explanation:

L=alpha*IL=αI
"(L:torque ;alpha :angular acceleration ;I:the moment of inertia)"(L:torque ;alpha :angular acceleration ;I:the moment of inertia)
alpha=(Delta omega)/(Delta t)α=ΔωΔt
alpha=(2*pi*Delta f)/(Delta t)α=2πΔfΔt
alpha=(2*pi*9)/6α=2π96
alpha=3*piα=3π
I=1/12*m*l^2 " for a rod spinning around its center"I=112ml2 for a rod spinning around its center
I=1/12*2*5^2=25/6I=112252=256
L=3*pi*25/6L=3π256
L=(25*pi)/2L=25π2