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.
To answer these questions, we need to apply the conservation of angular momentum and the principle of conservation of linear momentum.
(a) Calculate the angular velocity of the merry-go-round:
Angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The initial angular momentum before the boy starts running is zero because the merry-go-round is initially at rest. When the boy starts running, he exerts a torque on the merry-go-round, causing it to start spinning. Due to conservation of angular momentum, the final angular momentum will be equal to the initial angular momentum.
The initial angular momentum L_initial = 0, and the final angular momentum L_final = Iω.
Equating the two, we have:
0 = Iω
Rearranging the equation, we get:
ω = 0 / I
= 0 rad/s
Therefore, the angular velocity of the merry-go-round is 0 rad/s.
(b) Calculate the speed of the boy relative to the surface of the merry-go-round:
The speed of the boy relative to the surface of the merry-go-round can be found using the principle of conservation of linear momentum.
The initial linear momentum of the boy and the merry-go-round is zero since both are at rest. When the boy starts running, his linear momentum changes, but the total linear momentum of the system (the boy plus the merry-go-round) remains constant.
The linear momentum, p = m*v, where p is the linear momentum, m is the mass, and v is the velocity of an object.
Let the final velocity of the boy relative to the surface of the merry-go-round be v'.
Since the mass of the boy m_boy = 27 kg and his speed relative to the ground is v_boy = 0.6 m/s, the initial linear momentum of the boy m_boy*v_boy is given by 27 kg * 0.6 m/s.
Using conservation of linear momentum, the final linear momentum of the system will be zero, since the initial linear momentum was zero.
To get the final linear momentum, we can find the final velocity of the merry-go-round v_mg by using the formula:
m_boy * v_boy = -m_mg * v_mg
Where m_mg is the mass of the merry-go-round, which we'll calculate using the moment of inertia I and the radius r.
The formula for the moment of inertia of a disc is I = (1/2) * m_mg * r^2. Rearranging the equation, we get:
m_mg = 2 * I / r^2
Substituting the values, we have:
m_mg = 2 * 238 kg m^2 / (4 m)^2
Now, we can use this value to find the final velocity of the merry-go-round:
27 kg * 0.6 m/s = - (2 * 238 kg m^2 / (4 m)^2) * v_mg
Simplifying the equation, we get:
3.24 kg m/s = - (119 kg) * v_mg
Dividing both sides of the equation by -119 kg gives us:
v_mg = 3.24 kg m/s / 119 kg
≈ -0.0272 m/s
Therefore, the speed of the boy relative to the surface of the merry-go-round is approximately 0.0272 m/s in the opposite direction.