Answer :
198J minimum work is required to get the cylinder rolling without slipping at a rotational speed of 20 s−1
Work corresponds to a change in energy. So, the work required to make this cylinder roll is equal to its final kinetic energy minus its initial kinetic energy.
W=Kf−Ki
We can assume that this cylinder begins from rest, in which case Ki=0.
W=Kf
What is this cylinder's new kinetic energy? If an object is rolling, it is undergoing both rotational and translational motion. So, the cylinder's kinetic energy is equal to its translational kinetic energy plus rotational kinetic energy.
W=1/2mv2+1/2Iω²
We have values for the cylinder's mass and its rotational speed. What about its translational speed? We can use the conversion formula
v=rω and substitute this into our equation.
W=1/2m(rω)²+1/2Iω²
W=1/2mr²ω²+1/2Iω²
What about the moment of inertia? We can look up in a table that a cylinder has moment of inertia 1/2mr². Plugging this in, our equation becomes:
W=1/2mr²ω²+1/2(1/2mr²)ω²
W=1/2mr²ω²+1/4mr²ω²
W=3/4mr²ω²
Now, let's plug in values for all these variables and get our answer.
W=34(66)(0.1)2(20)2
W=198J
A solid 66 kg cylinder has a radius of 0.10 m. What minimum work is required to get the cylinder rolling without slipping at a rotational speed of 20 s−1?
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