Sample problem 9-2 Three particles of Fig 9-9 masses m,=2.3kg, y(m) m,=3.2kg and m=1.5kg are F=4.5N connected by thin rods of negligible mass, so that they lie at the vertices of 3-4-5 right triangle in the x-y 4X(m) dane
Sample problem 9-2 Three particles of masses =2.3kg, =3.2kg and =1.5kg are connected by thin rods of negligible mass, so that they lie at the vertices of 3-4-5 right triangle in the x-y plane. Fig 9-9 3 4 m1 m2 m3 X (m) y (m) c 1 r 2 r 3 r → F =4.5N 30 m2 m3 m1
Question: (a) find the rotational inertia about each of the three axes perpendicular to the x-y plane and passing through one of the particles. (b)a force of magnitude 4.5N is applied to m, in the xy plane and makes an angles of 300 with the horizontal, find the angular acceleration about oz axis (c) Find the rotational inertial about an axis perpendicular to the xy plane and passing through the center of mass of the system
Question: (a) Find the rotational inertia about each of the three axes perpendicular to the x-y plane and passing through one of the particles. (b) A force of magnitude 4.5N is applied to m2 in the xy plane and makes an angles of 300 with the horizontal. Find the angular acceleration about oz axis. (c) Find the rotational inertial about an axis perpendicular to the xy plane and passing through the center of mass of the system
Solution (a) consider first axis through =(23kg)(Om)2+(32kg).(3.0m)2+(15kg)(4m) 53kg.m Similarly for the axis through m2, we have 2=(23kg)(30m)2+(32kg)(Om)2+(15kg)(50m) 58kg·m y(m) For the axis through m F=45N 30 3=117kg:m2 C h
Solution: (a) consider first axis through Similarly for the axis through , we have For the axis through 2 2 2 2 2 1 53 (2.3 ) (0 ) (3.2 ) (3.0 ) (1.5 ) (4 ) k g m k g m k g m k g m I m rn n = = + + = 2 2 2 2 2 58 (2.3 ) (3.0 ) (3.2 ) (0 ) (1.5 ) (5.0 ) k g m I k g m k g m k g m = = + + 2 I 3 = 117k g m m1 m2 m3 3 1 4 m m2 m3 X (m) y (m) c 1 r 2 r 3 r → F =4.5N 30
(b Using Eq(9-2) Z=F=-(3m):(4.5N)cos30 (3m)·(4.5N) 53kg·m cOS 30 (cRotational inertial about an axis passing through the cm n n =0.86m =1.37m
(b) Using Eq(9-2) (c) Rotational inertial about an axis passing through the cm. z = = − rF m N ⊥ (3 ) (4.5 )cos30 2 1 (3 ) (4.5 ) cos 30 53 z z m N I kg m = = − m m m y y n n n cm = =1.37 m m m x x n n n cm = = 0.86
+ymn=2.62m 2 )2=3.40m 2 =11.74 2 ∑m2 =(23kg)(262m2)+(32kg)(340m2)+(15kg)(174m2) = 35kg. m cm111213
2 2 2 2 r1 = xcm + ycm = 2.62m 2 2 2 2 2 r2 = xcm + (y − ycm ) = 3.40m 2 2 2 3 2 r3 = (x − xcm ) + ycm =11.74m 2 2 2 2 2 35 (2.3 ) (2.62 ) (3.2 ) (3.40 ) (1.5 ) (11.74 ) k g m k g m k g m k g m I m r cm n n = = + + = Icm<I1 ,I2 ,I3