Question

An Atwood machine consists of two masses, mA= 63 kg and mB = 71 kg , connected by a massless inelastic cord that passes over a pulley free to rotate (Figure 1). The pulley is a solid cylinder of radius R = 0.40 mm and mass 5.0 kg. [Hint: The tensions FTA and FTB are not equal.]

Acceleration of each mass is 0.57 m/s²

What % error would be made if the moment of inertia of the pulley is ignored?

Express your answer using two significant figures.

Answer 1

From the data (my >ma),the net force acting on ma & mois, m;& -Ti = mga 7 - m48 = ma Adding the equations from one to another,solve for a (mg-ma)g-(ma+my) a = 7; -7, (mg-ma) 8-(ma+mz) a = 1 R here ,Moment of inertia of the pulley is, MR2 2 Therefore, (mg-ma)g-ma+mg) a = MR2 2 R2 ma+m2+ (mg-ma)g (1) 2 → From the given data, acceleration of each mass is, Vo = 0.57m/s If Moment of inerti a of pulley ignored (I w = 0 & M = 0), equation (1) becomes, (m; -ma)! (ma+m) (71kg - 63ks) (9.8m/s) (63 kg +71kg) = 10.585m/s Perrcentage change in acceleration of the system can be calculated as follows: lao - a %a 20 0.57 m/s2 - 0.585 m/s2 0.57m/s2 2.6446% = 2.6% * → Two siginficant figures