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JOULES TO RADIANS PERSECOND PLUS

m 2)(1.51 rad/s) 2 = 568 J Thus the energy has increased by 94 J.Ĭ) The extra energy had to have come from somewhere, in this caseįrom stored (chemical) energy in the child's muscles - the child did work on the merry-go-round while walking in to the center.

m 2)(1.257 rad/s) 2 = 474 J the final rotational kinetic energy is K f = ½I f f 2 = ½(500 kg.

m 2) = 1.51 rad/secī) The initial rotational kinetic energy is K i = ½I i i 2 = ½(600 kg.

No rotational inertia in the final situation, so the I f

m 2 Since the child ends up at the center (r=0), she/he contributes.

Of the child located at the rim I i = 500 Kg

JOULES TO RADIANS PERSECOND PLUS

(T = 5 seconds) so the initial angular velocity is i = 2 /T = 1.257 rad/s The initial moment-of-inertia is that of the merry-go-round plus that I i i/ I f so all that is left to do is to find the initial angular velocityĪnd the initial and final moments of inertia. On the system of merry-go-round and child). Starts at the outside and walks inward instead of vice-versa.Īpply conservation of angular momentum (there are no net external torques

The P300 engine in the XE produces 300 PS at 5,500 rpm and 400 Nm of torque between 1,500 4,500. The child can be considered as a point mass.Ī) What is the new angular velocity of the merry-go-round? b) Does the total kinetic energy of the system increase orĭecrease? By how much? c) Where does this change in energy go to or come from? Solution:Ī) This is just like problem 10-49, except this time the child Read the latest contents about horsepower calculator nm rpm in Malaysia, Check out Latest Car News, Auto Launch Updates and Expert Views on Malaysia Car Industry at WapCar.

Merry-go-round (about an axis through its center) isĪ child of mass 25 kg, originally standing at the rim, walks in For example, an acceleration might be in terms of m/s2 or meters per second squared. A merry-go-round of radius 2 m is rotating about a frictionless pivot. Note that compound units are formed by using, / and.