In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding
ID: 2205539 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding block has a mass of m2 = 0.890 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is ?k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table
Explanation / Answer
Hi, The pulley is not massless. So, the tension will not be uniform throughout the rope although the rope mass is negligible. Now, consider the free body diagram of the mass m1 (I am unable to represent the figure here, However, I shall try to explain the concept) Let T1 be the tension pulling the mass m1, So, T1 will be acting upwards and m1*g where g is the acceleration due to gravity downwards. Let a be the acceleration of the system So, m1g-T1=m1a ---------------(1) Let T2 be the tension pulling the mass m2 As the block m2 is sliding on the horizontal surface, the friction force acts and it opposes the motion of the block m2 So, T2-(.25)m2g=m2a ------------(2) (where .25 is the coefficient of kinetic friction) Now, writing the equations of the pulley which is having mass of .35 kg, Torque(T)=(Moment of inertia(I))*(angular acceleration(@) => T1*R2-T2*R2=(m(R2^2-R1^2)/2)*@=(m(R2^2-R1^2)/2)*a/R2 ----(as @=a/R2) =>T1-T2=.0972a -------------(3) Solving the equations (1),(2) and (3), we get a=1.8 m/sec^2 Kindly let me know if you need any clarification in this regard...