A 27-kg boy stands 4 m from the center of a frictionless playground merry-go-round, which has a moment of inertia of 238 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.
(b) Calculate the speed of the boy relative to the surface of the merry-go-round.
(a) Total angular momentum remains zero as the merry-go-round and the boy move in opposite directions.
M*V*R - I*w = 0
Solve for w
w = MVR/I = 27*0.6*4/238 = 0.27 rad/s
(b) Relative velocity to platform below the boy = V - (-R*w) = V + R*w
To answer these questions, we need to apply the principles of rotational motion and conservation of angular momentum.
(a) To find the angular velocity of the merry-go-round, we can use the conservation of angular momentum. Angular momentum is given by the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, before the boy starts running, the merry-go-round is not rotating, so its angular momentum is zero.
When the boy starts running, he imparts angular momentum to the system. The angular momentum of the boy is given by Lboy = Iboy × ωboy, where Iboy is the moment of inertia of the boy and ωboy is the angular velocity of the boy relative to the ground.
Since the system is isolated and there is no external torque acting on it, the total angular momentum remains constant. Therefore, we can equate the initial and final angular momentum:
Lboy = -Lmerry-go-round (Note the negative sign as the merry-go-round rotates in the opposite direction to the boy's motion.)
Iboy × ωboy = -Imerry-go-round × ωmerry-go-round
Substituting the given values:
27 kg × (0.6 m/s) × (4 m) = -238 kg m^2 × ωmerry-go-round
Simplifying the equation:
7.2 kg m/s = -238 kg m^2 × ωmerry-go-round
Rearranging the equation to solve for ωmerry-go-round:
ωmerry-go-round = (7.2 kg m/s) / (-238 kg m^2)
ωmerry-go-round ≈ -0.030 m/s
So, the angular velocity of the merry-go-round is approximately -0.030 m/s. The negative sign indicates that the merry-go-round is rotating in the opposite direction to the boy's motion.
(b) To find the speed of the boy relative to the surface of the merry-go-round, we need to consider the concept of relative velocities. The velocity of the boy relative to the merry-go-round is the vector sum of their velocities, where both are treated as vectors.
The velocity of the boy relative to the merry-go-round, vboy-merry-go-round, is given by:
vboy-merry-go-round = vboy - vmerry-go-round
Where vboy is the velocity of the boy relative to the ground and vmerry-go-round is the velocity of the merry-go-round.
Given that the velocity of the boy relative to the ground is 0.6 m/s, and we have already calculated the angular velocity of the merry-go-round as -0.030 m/s, we can use the formula:
vmerry-go-round = R × ωmerry-go-round
Substituting the given values:
vmerry-go-round = (4 m) × (-0.030 m/s)
vmerry-go-round = -0.12 m/s
Now, calculate the velocity of the boy relative to the merry-go-round:
vboy-merry-go-round = (0.6 m/s) - (-0.12 m/s)
vboy-merry-go-round = 0.72 m/s
So, the speed of the boy relative to the surface of the merry-go-round is 0.72 m/s.