Question

3. The 0.5 kg pulley with moment of inertia lo 6.25 x 10 kg m2 is released from rest at time t -0. There is no slipping of the cord, which is wrapped over the pulley. Neglect the weight of the cord and find: a. ?, the angular acceleration of the pulley b. T~ T, the tensions on each side of the pulley c·?, the angular velocity at t-3 [s] 250 1P Toog

Answer 1

let mA = 250 g = 0.25 kg
mB = 100 g = 0.1 kg
R = 0.05 m

let alfa is the angular acceleration of the pulley and a is the acceleration of the bodies.

Net force acting on mA, FnetA = mA*g - TA

mA*a = mA*g - TA

==> TA = mA*g - mA*a -------(1)

Net force acting on mB, FnetB = TB - mB*g

mB*a = TB - mB*g

==> TB = mB*g + mB*a --------(2)

Net Torque acting on pulley, Tnet = I*alfa

(TA - TB)*R = I*a/R

TA - TB = I*a/R^2

mA*g - mA*a - mB*g - mB*a = I*a/R^2

(mA - mB)*g = a*(I/R^2 + mA + mB)

==> a = (mA - mB)*g/(I/R^2 + mA + mB)

= (0.25 - 0.1)*9.8/(6.25*10^-4/0.05^2 + 0.25 + 0.1)

= 2.45 m/s^2

alfa = a/R

= 2.45/0.05

= 49 rad/s^2 <<<<<<<-----------Answer

b) TA = mA*g - mA*a

= 0.25*9.8 - 0.25*2.45

= 1.8375 N <<<<<<<-----------Answer

TB = mB*g + mB*a

= 0.1*9.8 + 0.1*2.45

= 1.225 N <<<<<<<-----------Answer

c) w = wo + alfa*t

= 0 + 49*3

= 147 rad/s <<<<<<<-----------Answer