The minimum work required to get the cylinder rolling without slipping at a rotational speed of 20 s-1 is 1.04 kg-m2/s2.
The torque required to rotate a solid cylinder without slipping is given by:
τ = Iα
Where I is the moment of inertia of the cylinder, and α is its angular acceleration.
The moment of inertia of a solid cylinder is given by:
I = mr2
Where m is the mass of the cylinder, and r is its radius.
Plugging in the values, we get:
I = (52 kg)(0.1 m)2 = 0.052 kg-m2
The angular acceleration is given by:
α = ω/t
Where ω is the angular velocity and t is the time taken to reach the desired angular velocity.
Plugging in the values, we get:
α = (20 s-1)/t
Since we do not know the time taken to reach the desired angular velocity, we will assume that it is 1 second.
Plugging in the values, we get:
α = (20 s-1)/1 s = 20 rad/s2
Now, we can calculate the torque required to rotate the cylinder without slipping:
τ = Iα = (0.052 kg-m2)(20 rad/s2) = 1.04 kg-m2/s2
Finally, we can calculate the minimum work required to get the cylinder rolling without slipping:
W = τθ
Where θ is the angular displacement. Since we do not know the angular displacement, we will assume it to be 1 radian.
Plugging in the values, we get:
W = (1.04 kg-m2/s2)(1 rad) = 1.04 kg-m2/s2
Hence, the minimum work required to get the cylinder rolling without slipping at a rotational speed of 20 s-1 is 1.04 kg-m2/s2.'
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Complete question:
A solid 52-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?