1.Consider a thin 28 m rod pivoted at one end. A uniform density spherical objec
ID: 1973549 • Letter: 1
Question
1.Consider a thin 28 m rod pivoted at one end. A uniform density spherical object
(whose mass is 6 kg and radius is 7.6 m) is attached to the free end of the rod and the
moment of inertia of the rod about an end is Irod =1/3mL2 and the moment of inertia of the sphere about its center of mass is Isphere =2/5m2r2.
i. Determine the torque generated by the motor to bring the wheel to 9.12 rev/min.
Answer in units of N · m
ii. Determine the power needed to maintain the rotational speed at 9.12 rev/min.
Answer in units of W
2.A constant torque of 26.2 N · m is appliedto a grindstone whose moment of inertia is 0.115 kg · m2.
Using energy principles, and neglecting friction, nd the angular speed after the grindstone has made 11.5 rev, assuming it startedfrom rest.
Answer in units of rev/s
Explanation / Answer
the rod has also mass of 9 kg. It has center of mass at 7m from the pivot. So there are two mass points, one, the rod of 9 kg at 7m from the pivot and the other the the sphere of 9 kg at a distance 14m + 3.2m = 17.2m from the pivot. we have to find out the CM of the two body system. if it is at x m from the pivot, then (x - 7)m . 9kg = (17.2 - x) . 9kg . Hence x = 12.1m. The entire 9kg +9kg = 18 kg mass may be supposed to be concentrated at this point, 12.1m from the pivot. So the moment of inertia of the system having mass 18kg at a distance 12.1 m from the pivot is I = MR^2 = 18 kg. (12.1).(12.1) m^2 . Force on the system is the weight 18kg.9.8m/s^2 acting downwards. So the torque is L = (12.1m). 18kg.9.8m/s^2 sin 35^0 = I . angular acceleration.
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