Engineers are designing a system by which a falling mass m imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (?Fig. P9.62?). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s2. In the earth tests, when m is set to 15.0 kg and allowed to fall through 5.00 m, it gives 250.0 J of kinetic energy to the drum. (a) If the sys-tem is operated on Mars, through what distance would the 15.0-kg mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the 15.0-kg mass be moving on Mars just as the drum gained 250.0 J of kinetic energy?

Solution 68 P Step 1: Data given Mass on earth m = 15 kg Distance on earth de= 5.0 m Kinetic energy of the drum K = 250.0 J 2 Gravitational force on Mars gM = 3.71 m/s We need to find the distance of fall to get same amount of kinetic energy on Mars Considering the diagram given This is obtained using KE = PE K + K + mgh = 0 m drum f The kinetic energy of mass is given by Km= 1/2 mv 2 f The kinetic energy of drum is given by Kdrum = 1/2 I o2 Here = vo/R f o The ratio of K drum and K m is given by Kdrum 1/2 Io Km = 1/2 mv2 f Kdrum = I/mR 2 Km o Here we can see that the kinetic energy does not depend on final velocity v f Hence K Mars = K Earth = v Mars = v Earth Hence we have the distance as GE G × de= d m M Substituting the values we get 9.8 m/s 3.71 m/s× 5.0 m = 13.2 m Hence the distance from the Earth is 13.2 m