Suppose m= 600g (including mass of holder) is released from rest. The pulley has
ID: 2015887 • Letter: S
Question
Suppose m= 600g (including mass of holder) is released from rest. The pulley has diameter d=2.65 cm, and you measure the angular acceleration of the pulley equals to 0.675 rad/s^2. Assume: the system is frictionless, the string does not slip on the pulley. The string is always perpendicular to the diameter of the pulley.a) calculate the linear acceleration of m as it falls.
b) Calculate T, the tension in the string
c) Calculate the torque exerted by the string on the pulley
d) Calculate I, the moment of inertia of the system
Explanation / Answer
Given that mass of the body is = 600 g = 0.6 kg diameter of pulley is(d) = 2.65 cm =0.0265 m angular acceleration of pulley is () = 0.675 rad/ s2 (a) Linear acceleration of body (a) = r 2where r is radius of pulley = d / 2 = 0.0265 / 2 =0.01325 m Linear acceleration of body (a) = (0.01325) (0.675) 2 = 0.006037 m / s2
(b) Tension in the string is given by from free body diagram mg - T = ma T = m (g-a) T = 0.6(9.8 -0.006037) T = 5.8763 N (c) Torque exerted by string on pulley is given by = force exerted on pully by string x radius of pulley = m a x 0.01325 where force F = m a = 0.6 x 0.006037 x 0.01325 = 0.000047994 = 4.7994 x 10-5 N (d) Moment of inertia of system is (I) = m r2 = 0.6 (0.01325)2 =0.00010535 kg m2 =0.00010535 kg m2