Question

Problem 5. The Figure below show two views of a double pulley system (two pulley wheels attached and sharing an axis of rotation) with radii r = 10 cm and R = 30 cm. The pulleys are solid cylinders with mass 0.5 kg and 3.5 kg. A string is wrapped around the large pulley and attached to a mass m of 10 kg. A string is wrapped around the small pulley and attached to a 900 mass M of 20 kg. The system is released from rest. What is the angular acceleration (both magnitude and direction) of the double pulley? TEI I = 2m12 T=fir E x 7 V 98. IN V 196,20 Small a We(301) 45.5k) (30m) a 1962169 m)=165)60) a 7848=0 18.69=&

Answer 1

Let angular acceleration be alpha (counterclockwise as positive), tensions in left and right be T1 and T2,

By force equation on blocks: m1g - T1 = m1a = m1*alpha*R or 10*9.8-T1 = 10*alpha*0.30 or 98-T1=3*alpha

T2 - Mg = M*alpha*r or T2 -20*9.8 = 20*alpha*0.10 or T2 -196 = 2*alpha

Torque equation on pulley : T1R - T2r = [0.5*0.5*0.10^2 + 0.5*3.5*0.30^2]*alpha

  T1*0.30 - T2*0.10 = [0.5*0.5*0.10^2 + 0.5*3.5*0.30^2]*alpha

So, 3T1 - T2 = 1.6 alpha

   putting the values of T1, T2

   3*(98-3alpha) - 2 alpha -196 = 1.6 alpha

   alpha = 7.778 rad/s^2

   direction is counterclockwise