Question

m2 m1 Two blocks, m, = 1.0 kg and m2 = 0.25 kg, are connected with a very light rope (neglect its mass) over a pulley with mass M = 0.5 kg and radius R = 0.25 m and moment of inertia I = 1/2 MR as shown in the drawing. The coefficient of kinetic friction between mand the table, HK = 0.4. Find a, the angular acceleration of the pulley.

Answer 1

a> T2 T2 Me m2 = m = 1kg 0.25kg k = 0.4 •T mag Ti a t = 0.25 m "mig 'M=0.5kg R = my g I - IMO For mass m, mig To = m, a Ti = m, cg-a) For mass M2 I R= Mag. fx = M R = M mag ::. Ta-fx = mza = Ta = m2 a + fx T2 Ta = maat M mag = ma (a + 4 g) for pulley e torque is given by 12 て IL = Z - I a also て (T, -12) From ③ € 6, I a (T-T2) Ti-T2 = I a la Put values from ③ & in Υ → m, fg-a) ma cat g) I a I a ra I a tm, at mza. = mg-m,a - Mza-mak g = (m. q m2 M) g g (m, uma Mc) = a (11 + mi +m2) g (m. - maq I + m, + m2) But ra à x0.5 x 0.asp? 0.252 a = та | Mga 11 72 = 0.25 put all values in = a = 9.81(1 – 0.25% 0-4) 0.25 + 1 + 0.25 5.886 misa? (Ans) ". Angular acceleration of pulley 5.886 25 a = 23.544 rad /s? Cans) -