Question

Let the masses of blocks a and b be 3.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.300 kg⋅m2, and the radius of the wheel be 0.100 m . find the magnitude of linear acceleration of block a if there is no slipping between the cord and the surface of the wheel.

Answer 1

Answer: Let: Mâ‚ = mass of block A = 3.50 kg Mâ‚‚ = mass of block B = 2.00 kg I = moment of inertia of the wheel = 0.400 kgm² R = radius of the wheel = 0.120 m g = acceleration by gravity = 9.8 m/s² a = linear acceleration = ? α = angular acceleration = ? Tâ‚ = tension in the string on the side of A = ? Tâ‚‚ = tension in the string on the side of B = ? Constructing the equation: Mâ‚Ă—g - Tâ‚ = Mâ‚Ă—a . . . . . . . . . . equation 1 Tâ‚‚ - Mâ‚‚Ă—g = Mâ‚‚Ă—a . . . . . . . . . . equation 2 (Tâ‚ - Tâ‚‚)Ă—R = I×α. . . . . . . . . . . equation 3 α = a / R . . . . . . . . . . . . . . . . equation 4 Solve this 4Ă—4 system of equations to find: a = 0.442 m/s² < - - - - - - - - - - answer 1 and 2 (both the same) α = 3.68 rad/s² < - - - - - - - - - - answer 3 Tâ‚ = 32.8 N < - - - - - - - - - - - - answer 4 or 5 (depending on which of) Tâ‚‚ = 20.5 N < - - - - - - - - - - - - answer 4 or 5 (the blocks is left or right)