(b) The acceleration of the two blocks is a= < op 2OR (c) The tensions in the lower sections of the string is T=M(g-a)=M(g-) 2R 201 (d)The tensions in the lower sections of the string is T,=T M(g--2)--2 R 3. A disk with moment of inertia I, is rotating with initial angular speed oo; a second disk with moment of inertia I2 initially is not rotating(see Figure 5). The arrangement is much like a LP record ready to drop Direction of onto an unpowered, freely spinning turntable. The h second disk drops onto the first and friction between them brings them to a common angular speed. Find the common angular speed o Solution Let two disks be a system, the total torque about the axis is zero, applying the conservation of angular momentum, we have the common angular speed o L 1o+120=1o0→(l1+12)b=l 1+12 4. Two cylinder having radii R, and R2 and rotational inertias 1, and 12, respectively, are supported by axes perpendicular to the plane of E如时四h( l1 linder and is caused to rotate by the frictional force between the two. Eventually, slipping ceases, and the two cylinders rotate at constant rates in opposite directions. Find the final angular velocity h of the small cylinder in terms of 11, 12, R1, R2, and @o(Hint: Angular momentum is not conserved Apply the angular impulse equation to each cylinder) Solution: Apply the rotational counterpart of Newton's second law of motion t dt ∫R,r=102-0 we get 「R,t=1a-1mo O,R=O2R2 Solving the equations, we can get the final angular velocity a of the small cylinder is(b) The acceleration of the two blocks is 2 2 t R a θ = (c) The tensions in the lower sections of the string is ) 2 ( ) ( 1 2 t R T M g a M g θ = − = − (d) The tensions in the lower sections of the string is 2 1 2 2 2 ) 2 ( t I t R M g R I T T α θ θ = − = − − 3. A disk with moment of inertia I1 is rotating with initial angular speed ω0; a second disk with moment of inertia I2 initially is not rotating (see Figure 5). The arrangement is much like a LP record ready to drop onto an unpowered, freely spinning turntable. The second disk drops onto the first and friction between them brings them to a common angular speed. Find the common angular speed ω. Solution: Let two disks be a system, the total torque about the axis is zero, applying the conservation of angular momentum, we have the common angular speed ω. 0 1 2 1 1 2 1 0 1 2 1 0 ω ω ω ( )ω ω ω ω I I I L L I I I I I I f i + = ⇒ + = ⇒ + = ⇒ = r r r r r r 4. Two cylinder having radii R1 and R2 and rotational inertias I1 and I2, respectively, are supported by axes perpendicular to the plane of Fig.6. The large cylinder is initially rotating with angular velocity ω0. The small cylinder is moved to the fight until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually, slipping ceases, and the two cylinders rotate at constant rates in opposite directions. Find the final angular velocity ω2 of the small cylinder in terms of I1, I2, R1, R2, and ω0. (Hint: Angular momentum is not conserved. Apply the angular impulse equation to each cylinder.) Solution: Apply the rotational counterpart of Newton’s second law of motion t L total d d v v τ = , we get ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = − = − = − ∫ ∫ 1 1 2 2 1 1 1 1 0 2 2 2 d d 0 R R R f t I I R f t I ω ω ω ω ω Solving the equations, we can get the final angular velocity ω2 of the small cylinder is ω0 r I2 I1 Direction of spin Fig.5 I2 I1 R2 R1 ω0 Fig.6 + +