A baton twirler in a marching band competition grabs one end of her 1.2 kg, 1.0 meter long baton. She throws her baton into the air such that it rises to a height of 5.0 meters while spinning end over end at a rate of 3.5 revolutions per second. How much work did she do on the baton?

349 J

Explanation:

Length L of baton = 1.0 m

Mass m of baton = 1.2 kg

Weight W of baton = 1.2 kg x 9.81 m/ = 11.772 N

Height h reached = 5.0 m

Angular speed ω = 3.5 rev/s = 2π x 3.5 (rad/s) = 21.99 rad/s

Total work done on baton will be the work done in taking it to a height of 5.0 m and the kinetic energy with which the baton rolls.

Work done to bringing it to the height of 5.0 m = weight x height above ground

W x h = 11.772 x 5 = 58.86 J

Velocity v of spinning baton = ω x L = 21.99 x 1 = 21.99 m/s

Kinetic energy = m =

Total work done on baton = 58.86 + 290.14 = 349 J

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more