A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turn
ed from the relaxed state through a small angle theta, the torque exerted by the fiber on the disk is proportional to theta. Find an expression for the torsional constant k in terms of the moment of inertia I of the disk and the angular frequency w of small, free oscillations. The disk, when twisted and released, oscillates with a period T of 1.00 s. Find the torsional constant k of the fiber.
1 answer:
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Answer:
Explanation:
speed of motor (N)=1500 rpm
power=4 hp = =2.9828 KW
service factor(k)= 2.75
now,
torque rating
Answer:
25 mm = 0.984252 inches
Explanation:
Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:
<u>1 mm = 1/25.4 inches</u>
From the question, we have to convert 25 mm into inches
Thus,
<u>25 mm = (1/25.4)*25 inches</u>
So,
Thus, solving we get:
<u>25 mm = 0.984252 inches</u>