Question

A 15.0-kg object and a m^2 =10.0-kg object are joined by a cord that passes over a pulley with a radius of R =10.0 cm and a mass of M = 3.00 kg. The cord has a negligible mass and does not slip on the pulley. The pulley rotates on its axis without friction. The objects are released from rest when they are 3.00m apart and are free to fall. Ignore air resistance. Treat the pulley as a uniform disk, and determine the speeds of the two objects as they pass each other.

Answer 1

Step-by-step explanation:

Let the common acceleration of the system of mass pulley be a .

For motion of 15 kg which is going down

15g - T₁ = 15a , T₁ is tension in the spring . ( motion of 15 kg )

T₂ - 3g = 3a , T₂ is tension in the spring .( motion of 3 kg )

(T₁ - T₂ ) x.10 = I α , I is moment of inertia of disc , α is angular acceleration ( rotational motion of pulley )

(T₁ - T₂ ) x.10 = 1/2 x 3 x .10² x( a / .10 ) ( linear accn = angular accn x radius )

(T₁ - T₂ ) = 3/2 a

adding previous equation

T₂ - T₁ + 12g = 18a

- 3/2 a + 12g = 18a

12g = 19.5 a

a = 6.03 m / s²

relative acceleration between the two = 12.06 m /s² ( they are coming towards each other )

distance = 3m

s = ut + 1/2 at²

3 = 1/2 x 12.06 t²

t = .71 s