A 23-kg boy stands 3 m from the center of a frictionless playground merry-go-round, which has a moment of inertia of 180 kg m2. The boy begins to run in a circular path with a speed of 0.6 m/s relative to the ground.
(a) Calculate the angular velocity of the merry-go-round in rad/s^2.
(b) Calculate the speed of the boy relative to the surface of the merry-go-round in m/s.
angular momentum of system remains zero
I w = m v r
180 kg m^2 * w = 23 kg * .6 m/s * 3 m
so
w = .23 radians/second
v merry = .23*3 = .69 m/s away from boy
so .6 + .69 = 1.29 m/s
Thank you!
To solve this problem, we can use the principle of conservation of angular momentum. The total angular momentum of the system remains constant, both before and after the boy starts running. The initial angular momentum of the system is zero since the boy is stationary. The final angular momentum of the system can be calculated as the sum of the angular momentum of the boy and the merry-go-round.
(a) To calculate the angular velocity of the merry-go-round, we can use the equation:
initial angular momentum = final angular momentum
Since the initial angular momentum is zero, we have:
0 = (moment of inertia of the merry-go-round) * (final angular velocity of the merry-go-round)
Rearranging the equation, we can solve for the final angular velocity:
final angular velocity of the merry-go-round = 0 / (moment of inertia of the merry-go-round)
final angular velocity of the merry-go-round = 0 rad/s
Therefore, the angular velocity of the merry-go-round is 0 rad/s.
(b) To calculate the speed of the boy relative to the surface of the merry-go-round, we can use the equation:
final angular velocity of the boy = (distance from the center) * (final angular velocity of the merry-go-round)
Plugging in the values:
final angular velocity of the boy = (3 m) * (0 rad/s)
final angular velocity of the boy = 0 rad/s
Therefore, the speed of the boy relative to the surface of the merry-go-round is 0 m/s.
To calculate the angular velocity of the merry-go-round, we can use the principle of conservation of angular momentum. The initial angular momentum of the system consisting of the boy and the merry-go-round should be equal to the final angular momentum when the boy begins to run. The formula for angular momentum is given by:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
(a) Initially, the system is at rest, so the initial angular momentum is zero. When the boy begins to run, he creates an angular momentum. We can calculate it using the formula:
Lboy = I * ωboy
where Lboy is the angular momentum created by the boy.
Since the system's angular momentum remains conserved, we can equate the two angular momenta:
Lboy = L
I * ωboy = 0
We can solve this equation for ωboy:
ωboy = 0 / I
Since the moment of inertia I ≠ 0, the angular velocity of the merry-go-round will remain zero because there is no external torque acting on it.
(b) The speed of the boy relative to the surface of the merry-go-round can be calculated using the formula:
vboy = v - R * ω
where vboy is the speed of the boy relative to the surface of the merry-go-round, v is the speed of the boy relative to the ground, R is the distance from the center of the merry-go-round, and ω is the angular velocity of the merry-go-round (which is zero in this case).
Plugging in the given values:
vboy = 0.6 m/s - 3 m * 0
Since the angular velocity is zero, the speed of the boy relative to the surface of the merry-go-round is equal to his speed relative to the ground:
vboy = 0.6 m/s.
Therefore, the speed of the boy relative to the surface of the merry-go-round is 0.6 m/s.