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?
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.
To answer these questions, we'll need to apply the principles of rotational motion and conservation of angular momentum. Let's tackle each question step by step:
(a) To determine the final angular velocity of the merry-go-round after the child catches the ball, we can use the conservation of angular momentum. The initial angular momentum is zero since the merry-go-round is initially at rest. The final angular momentum can be calculated using the equation:
Initial Angular Momentum = Final Angular Momentum
The initial angular momentum is given by the moment of inertia of the merry-go-round multiplied by its initial angular velocity, which is zero:
0 = 100 kg * R^2 * 0
The final angular momentum is given by the moment of inertia of the merry-go-round plus the moment of inertia of the child (a point particle located at the edge):
Final Angular Momentum = (100 kg * R^2) * ω
where ω is the final angular velocity of the merry-go-round.
Since the angular momentum is conserved, we can solve for ω:
0 = (100 kg * R^2) * ω
ω = 0 rad/s
Therefore, the final angular velocity of the merry-go-round is 0 rad/s.
(b) The work done on the merry-go-round can be calculated using the equation:
Work = Change in Kinetic Energy
Since the initial and final angular velocities are both zero, there is no change in kinetic energy. Therefore, the work done on the merry-go-round is zero.
(c) When the work done on an object is zero, it means there is no change in energy. So, the work-energy diagram for the system consisting of only the merry-go-round would show a flat line, indicating that there is no change in energy.
(d) The merry-go-round does not do any work on the ball or the child. This is because work is defined as the transfer of energy through the application of a force over a distance. In this case, no external force is acting on the merry-go-round, so it does not do any work.
(e) The centripetal force acting on the ball after the child catches it can be calculated using the equation for centripetal force:
Centripetal Force = mass * velocity^2 / radius
Given: mass = 0.62 kg, velocity = 20 m/s, and radius = 1.5 m, we can plug in these values to calculate the centripetal force.
To solve this problem, we can use the principles of conservation of angular momentum and energy.
(a) The initial angular momentum of the system is 0 since the merry-go-round is at rest and the child is not rotating. When the child catches the ball, they will start rotating with the merry-go-round due to the conservation of angular momentum. The angular momentum of the system remains constant.
The angular momentum of the system is given by L = Iω, where I is the moment of inertia and ω is the angular velocity. The moment of inertia of the merry-go-round is M = 100 kg * 1.5 m^2 = 150 kg*m^2.
The angular momentum of the child and the ball can be calculated as follows:
L_child = m_child * r * v_child, where m_child is the mass of the child, r is the radius of the merry-go-round, and v_child is the velocity of the child.
L_child = (40 kg) * (1.5 m) * 0 = 0
L_ball = m_ball * r * v_ball, where m_ball is the mass of the ball and v_ball is the velocity of the ball.
L_ball = (0.62 kg) * (1.5 m) * 20 m/s = 18.6 kg*m^2/s
The total angular momentum after the child catches the ball is the sum of the angular momenta of the child and the ball:
L_total = L_child + L_ball = 0 + 18.6 kg*m^2/s = 18.6 kg*m^2/s
The final angular velocity of the merry-go-round can be found using the equation L = Iω:
L_total = I * ω
ω = L_total / I = 18.6 kg*m^2/s / 150 kg*m^2 = 0.124 rad/s
Therefore, the final angular velocity of the merry-go-round after the child catches the ball is 0.124 rad/s.
(b) The work done on the merry-go-round can be calculated using the equation:
Work = ΔKE + ΔPE
Since there is no change in potential energy for the merry-go-round, ΔPE = 0.
The change in kinetic energy (ΔKE) is equal to the work done on the merry-go-round. The initial kinetic energy of the merry-go-round is 0 since it is at rest, and the final kinetic energy is given by:
KE_final = (1/2) * I * ω^2 = (1/2) * (150 kg*m^2) * (0.124 rad/s)^2 = 0.918 J
Therefore, the work done on the merry-go-round is 0.918 J. This work increases the rotational kinetic energy of the merry-go-round.
(c) The work-energy diagram for the merry-go-round system consists of only the merry-go-round and shows that the work done on the system increases its kinetic energy. As mentioned earlier, there is no change in potential energy, so the diagram would show an arrow pointing only towards the kinetic energy component.
(d) The merry-go-round does not do any work on the ball or the child since there is no tangential force acting on them. The work done on the ball and the child is zero.
(e) The centripetal force acting on the ball after the child catches it is provided by the tension in the child's hand holding the ball. The centripetal force is given by:
F_c = (m_ball * v_ball^2) / r = (0.62 kg) * (20 m/s)^2 / 1.5 m = 165.33 N
Therefore, the centripetal force acting on the ball after the child catches it is 165.33 N.