A 0.7 kg mass at (x, y) = (20 cm,20 cm) and a 2.5 kg mass at (20 cm,100 cm) are
ID: 1681712 • Letter: A
Question
A 0.7 kg mass at (x, y) = (20 cm,20 cm) and a 2.5 kg mass at (20 cm,100 cm) are connected by a massless, rigid rod. They rotate aboutthe center of mass. (a) What are the coordinates of the center ofmass?( 1 cm, 2 cm)
(b) What is the moment of inertia about the center of mass?
3 kg·m2
(c) What constant torque will cause an angular velocity of 6.25rad/s at the end of 3.0 s, starting from rest?
4 N·m
(d) At what angle, with respect to the axis of the rod, should2.0 N forces be applied to each massto give the torque you found in part c?
5°
(a) What are the coordinates of the center ofmass?
( 1 cm, 2 cm)
(b) What is the moment of inertia about the center of mass?
3 kg·m2
(c) What constant torque will cause an angular velocity of 6.25rad/s at the end of 3.0 s, starting from rest?
4 N·m
(d) At what angle, with respect to the axis of the rod, should2.0 N forces be applied to each massto give the torque you found in part c?
5°
Explanation / Answer
(a)Let m1 = 0.7 kg and (x, y) = (20 cm, 20cm). Similarly,let m2 = 2.5 kg and(x1,y1) = (20 cm, 100 cm) The coordinates of the center of mass is (X,Y) = [(m1x +m2x1/m1 +m2),(m1y +m2y1/m1 + m2)] (b)The moment of inertia about the center of mass is I = (mL2/12) where m = (m1 + m2/2) and L2= (X - 20)2 + (Y - 20)2 (c)Let the constant torque be We know from the relation w = wo + t or = (w - wo/t) where w = 6.25 rad/s,wo = 0 and t = 3.0 s The constant torque is = I * (d)Let the angle be We know that = F * L * sin where F = 2.0 N or sin = (/F * L) or = sin-1(/F * L)Related Questions
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