Question

Consider a system containing a solid cylinder of mass 10 kg and diameter 0.5 m, a thin cylindrical shell of mass 2 kg and diameter 0.3 m, and a tbin spherical shell of mass 5 kg and radius 0.25 m arranged as shown in the image anove and all connected by a massless thin rod. The center of each object is 1 m apart in the system is free to rotate about an access 1 m to the left of the center of a solid cylinder. See image.

Just a Moment oom -1.00m om Tokg 2kg 5kg 00.30m- - RO.25m - $0.50m - Consider a system containing a solid cylinder of mass 10.0kg and diameter 500m, a thin cylindrical shell of mass 2.00kg and diameter 300m, and a thin spherical shell of mass 5kg and radius 0.250m arranged as shown in the image above and all connected by a massless thin rod. The center of each object is 1.00m apart and the system is free to rotate about an axis 1.00m to the left of the center of the solid cylinder. See image. I= mira A) Find each of the following moments of inertia Obiect Axis Moment of Inertia Solid Cylinder its own Center of Mass 5.0 Valm Cylindrical Shell Rotational Auxis oole kalmt Spherical Shell Its own Center of Mass 0.313 m2 Complete System Rotational Axis 10,938 kalma B) A torque of 5.00 m N is applied for a period of 8s. What is the new angular momentum? C) The spherical shell is now released from the rest of the system. What is the angular speed of the remaining system containing the cylinder and cylindrical shell? D) What is the speed of the spherical shell the moment after it is released? - Iw

Answer 1

A) solid cylinder about its own axis through center of moss Icramp? = x 10 kg x(0.5.00m) ? I c = 11.25 kg m² Cylindrical shell about rotational axis a Fart about it's central axis Igc = Mr² about rotation axis IR = Inc & Md² d distance between i I2R = Mr?+ md² l d=1.00m +1.com = 2 kg (0.30m)2 + 12.00m)2] IR = 18.18 kg m² the two axis = 2000 m Spherical shell about axis through its own center of mass Isc = 2 / Mr2 = 2 / 3 x 5 kg x (0.25 m2 Ige = 4 10.2083 kgm] Complete system about Rotatational axis solid cylinder about Rotational axis. FIR = Ile + Md² = 1.25kgm² + 10kgx (1room? FR = 11:25 Kgm²

Spherical shell about rotational axis I3R = Isc + md² - 0.2083 + 5kg x(3.00m)? = 45,2083 Kg ma whole system about rotational axis Ij = FR + I2R + IBR = 11.25 kgm² + 8.18 kgm2 + 45:2083 Kgm² Ij = 164.6383 kg m² Br torque t= 5:00 N time t =85 change in angular momentum = torque x time Li-to = 2t . Lo=0 (initial angular momentum) - - L = Tt. = 500 m®N x 85 LA = 40.0 km N.ms Li = Ii Wi Cinitial momentum before ) (Shell is released Wir ha

W = 40.0 Nms 45.2083 kgm? Di = 0.8848 rodds (initial angular speed of system) As per the nature of question let assume that spherical I spheshell after released doesn't have rotational motion have only linear speed given by wild where d = 3:00 distance of center of spherical shell from the rotational axis.' Conservation of Angular momentum Lf = Li (Tip + I2R) Wy + Mss (Wid) d = i Mss = mass of spherical shell. 1. Mss Wid). I angulos momentum of spherical shell, I just after the release about rotational axis (IR + I2R) 7 moment of inestia of remaining system 1. Wf = final angular speed (FIR + F 2R) of = Angular momentum of remaining system Ws = Li - Mss d coi -- (40:0 Nms – 516g xß.00m)? IRTIR X 0.8848 11.25 kym² + 8 18 kgm?

2 d) wf = 10.00947 rad/s speed of spherical shell = wid = 0.8848 radls x 3.00m = /2.6544 m/s 7