Using the equation t=FXF=la, we have rF9.88×10 11.14x10~3(04961+030512)=42.9871+20643y3 42.987t+26.433t If the pulley was initially at rest, thus its angular speed after 3. 60s is O=(42.987+264332)d=21492+88r1=6896rads 2. Two identical blocks, each of mass M, are connected by a 72 ight string over a frictionless pulley of radius R and rotational inertia I( Fig 4). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and sliding block. When this system is released, it is found that the pulley turns through an angle 8 in time t and the acceleration of the blocks is constant.(a)What is the angular acceleration of the pulley?(b)What is the acceleration of the wo blocks?(c)What are the tensions in the upper and lower sections of the string? All answer are to be expressed in terms of M,,R, e, g, and t Solution: Sketch the forces diagram of the system shown in figure N Mg m8↓T1 Apply the Newton's second law and the counterpart of Newtons law for a spinning rigid body, we Mg -l= Ma (l-TR=la a=aR Solving them, we get a)The angular acceleration of the pulley is aUsing the equation τ α v v v v = r × F = I , we have 2 2 3 2 (0.496 0.305 ) 42.987 26.433 1.14 10 9.88 10 t t t t I rF z + = + × × = = − − α For dt dω α = , so 2 42.987 26.433 d d t t t = = + ω α , If the pulley was initially at rest, thus its angular speed after 3.60s is (42.987 26.433 )d 21.494 8.811 689.6 rad/s 3.6 0 2 3 3.6 0 2 = + = + = ∫ ω t t t t t 2. Two identical blocks, each of mass M, are connected by a light string over a frictionless pulley of radius R and rotational inertia I (Fig.4). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and sliding block. When this system is released, it is found that the pulley turns through an angle θ in time t and the acceleration of the blocks is constant. (a) What is the angular acceleration of the pulley? (b) What is the acceleration of the two blocks? (c) What are the tensions in the upper and lower sections of the string? All answer are to be expressed in terms of M, I, R, θ, g, and t. Solution: Sketch the forces diagram of the system shown in figure. Apply the Newton’s second law and the counterpart of Newton’s law for a spinning rigid body, we have ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎨ ⎧ = = − = − = 2 1 2 1 2 1 ( ) t a R T T R I Mg T Ma θ α α α Solving them, we get (a) The angular acceleration of the pulley is 2 2 t θ α = . M M T2 T1 I Fig.4 Mg f T1 T2 T1 T2 N Mg N mg a v