The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass mp=5.33 kg and radius rp=0.150 m. The hanging masses are mL=20.5 kg and mR=11.7 kg. Two masses are connected by a rope that passes over a pulley. The mass labeled M subscript L hangs directly beneath the left edge of the pulley. The mass labeled M subscript R is smaller than M sub L and hangs directly beneath the right edge of the pulley. M sub R is lower than M sub L in its initial position. Calculate the magnitude of the masses' acceleration a and the tension in the left and right ends of the rope, TL and TR , respectively.

Respuesta :

Answer:

Explanation:

Let the common  acceleration be a .

Considering motion of mL

20.5g - TL = 20.5 a

Considering motion of mR

TR - 11.7g = 11.7 a

TR-TL + 20.5 g- 11.7 g = (20.5 + 11.7 ) a

TL-TR = 8.8 g - 32.2 a

Considering motion of wheel

( TL - TR ) x rp = 1/2 x mp x rp² x α where α is angular acceleration of wheel

=  ( TL - TR )  = 1/2 x mp x a      (  a = α x rp )

Putting the values from equation above

8.8g - 32.2a = .5 mp a

8.8g = 32.2 a + .5 x 5.33 a

8.8 g = 34.865 a

a = 2.47 m /s²

20.5 g - TL = 20.5 a

TL = 20.5 x 9.8 - 20.5 x 2.47

= 150.26 N

TR - 11.7g = 11.7 a

TR - 11.7 x 9.8  = 11.7 x 2.47

TR = 143.56 N