a. The linear momentum of the ball
(m(o) – 0.64 kg), v =20 m/s) is
p = m(o)•v .
The angular momentum is
L = m(o)•v•R .
On the other hand, the angular momentum is
L =I•ω,
where the total moment of inertia (merry-go-round + the child) is
I = mR^2 +MR^2/2.
m(o)•v•R =( mR^2 +MR^2/2) •ω,
ω = m(o)•v/(R•(m+M/2)) = 0.092 rad/s.
b.The work is the change of the rotational kinetic energy
W = ΔKE = I•ω^2/2 = (mR^2 +MR^2/2) •ω^2/2 =202.5•(0.092)^2/2 = 34.28 J.
c. -
d. -
e.F =m•a=m• ω^2•R =0.062•(0.092)^2•1.5 = 7.87•10^-3 N.
A 40 kg child is standing on the edge of a merry-go-round in a playground. Before they were deemed too dangerous, these were quite common. They were just huge rotating platforms you could sit on while someone spun you around in uniform circular motion until you screamed in terror (or possibly got sick). At any rate, let us assume the merry- go-round is a rotating disk of radius R = 1.5m and inertia M = 100kg and the child is a point particle located at the edge of the disk. The merry-go-round is initially at rest. The child’s brother tosses him a basketball from across the playground. Assume the basketball has a mass of 0.62kg and is thrown with a velocity of 20m/s perpendicular to the radius of the merry-go-round at the point where the child catches it.
(a)[7 pt(s) ]Assume that there is no friction in the axle of the merry-go-round, so it can rotate freely. Assume that there is a lot of friction between the child and the merry-go-round, so that he does not slide as he catches the ball. What is the final angular velocity of the merry-go-round after the child catches the ball?
(b)[4 pt(s) ]How much work is done on the merry-go-round? What does this work?
(c)[3 pt(s) ]Draw the work-energy diagram to express the change in energy of the system consisting of only the merry-go-round.
(d)[4 pt(s) ]Does the merry-go-round do any work on the ball? On the child? How much?
(e)[3 pt(s) ]What is the centripetal force acting on the ball after the child catches it?
1 answer
1 answer