A merry-go-round is a common piece of playground equipment. A 2.4 m diameter merry-go-round with a mass of 300 kg is spinning at 22 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 25 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?
Well, well, well, it seems like John wants to join the merry-go-round party! Let's calculate the angular velocity after his daring jump.
First, we need to find the initial angular momentum of the merry-go-round before John jumps on. Angular momentum is given by the equation:
L = I * ω,
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid disk is given by:
I = (1/2) * m * r^2,
where m is the mass of the merry-go-round and r is the radius.
Plugging in the values, we get:
I = (1/2) * 300 kg * (2.4 m/2)^2 = 180 kg*m^2.
Now, let's find the initial angular velocity:
22 rpm = ω * (2π rad/1 min),
Converting revolutions per minute to radians per second (because we love consistency):
ω = (22 rpm) * (2π rad/1 min) * (1 min/60 s) ≈ 7.28 rad/s.
Now, let's get ready for John's entrance!
When John jumps onto the merry-go-round, angular momentum is conserved, which means the initial angular momentum must be equal to the final angular momentum.
Since John jumps onto the outer edge of the merry-go-round, the moment of inertia of the system becomes:
I_final = I_merryGoRound + I_John,
where I_merryGoRound is the moment of inertia of the merry-go-round, and I_John is John's moment of inertia.
The moment of inertia for John can be approximated as the mass of John times the square of the distance from the center of the merry-go-round to where John lands (the outer edge).
I_John = m_John * r^2 = 25 kg * (2.4 m/2)^2 = 90 kg*m^2.
Thus, the final moment of inertia of the system is:
I_final = 180 kg*m^2 + 90 kg*m^2 = 270 kg*m^2.
So, the final angular velocity can be found by rearranging the equation for angular momentum:
L_final = I_final * ω_final,
ω_final = L_final / I_final,
ω_final = (I_initial * ω_initial) / I_final,
where L_initial is the initial angular momentum of the system, which is equal to I_initial * ω_initial (before John jumps on).
Plugging in the values, we find:
ω_final = (180 kg*m^2 * 7.28 rad/s) / 270 kg*m^2 ≈ 4.86 rad/s.
Now, let's convert the angular velocity to rpm:
ω_final_rpm = (ω_final * 1 min/2π rad) * (1 rev/1 cycle) * (60 s/1 min),
ω_final_rpm ≈ (4.86 rad/s * 1 min/2π rad) * (1 rev/1 cycle) * (60 s/1 min),
ω_final_rpm ≈ 4.65 rpm.
So, after John's epic jump, the merry-go-round will be spinning at approximately 4.65 rpm.
Now, isn't that a wonderful twist to the story?
To find the merry-go-round's angular velocity after John jumps on, we can use the principle of conservation of angular momentum.
1. First, let's calculate the initial angular momentum of the merry-go-round before John jumps on. The formula for angular momentum is L = Iω, where L is angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia of a circular object like a merry-go-round is given by the formula I = (1/2)mr², where m is the mass and r is the radius.
Given:
Diameter of the merry-go-round (d) = 2.4 m
Radius (r) = d/2 = 2.4/2 = 1.2 m
Mass of the merry-go-round (M) = 300 kg
Angular velocity (ω) = 22 rpm
Moment of Inertia (I) = (1/2) M r²
= (1/2) * 300 * (1.2)^2
= 216 kg m²
Angular momentum (L) = I * ω
= 216 * 22
= 4752 kg m²/s
2. Next, let's calculate John's angular momentum before he jumps on. Since he is initially running tangent to the merry-go-round, his angular momentum is 0 because his distance from the axis of rotation is 0.
3. According to the principle of conservation of angular momentum, the total angular momentum before John jumps on should be equal to the total angular momentum after he jumps on.
The total angular momentum after John jumps on can be calculated using the formula L' = I'ω'.
Here, I' is the moment of inertia of the new system (merry-go-round + John) after John jumps on, and ω' is the final angular velocity we need to find.
4. Let's calculate the moment of inertia of the new system after John jumps on.
Moment of Inertia of the merry-go-round (I_m) = 216 kg m²
Mass of John (m_J) = 25 kg
The moment of inertia of a point mass is given by I_J = m_J * r_J², where r_J is the distance of John from the axis of rotation.
Since John jumps on the outer edge of the merry-go-round, r_J = r (radius of the merry-go-round).
Moment of Inertia of John (I_J) = m_J * r_J²
= 25 * (1.2)^2
= 36 kg m²
Total Moment of Inertia after John jumps on (I') = I_m + I_J
= 216 + 36
= 252 kg m²
5. Now, we can use the conservation of angular momentum to find the final angular velocity (ω') of the merry-go-round after John jumps on.
L = L'
Iω = I'ω'
216 * 22 = 252 * ω'
4752 = 252 * ω'
ω' = 4752 / 252
ω' = 18.857 rpm
Therefore, the merry-go-round's angular velocity, after John jumps on, is approximately 18.857 rpm.
To solve this problem, we need to apply the principle of conservation of angular momentum. The angular momentum of an object is the product of its moment of inertia and its angular velocity.
First, let's find the initial angular momentum of the merry-go-round before John jumps on. The moment of inertia (I) of a solid disk rotating about its center is given by the formula: I = 0.5 * m * r^2, where m is the mass of the object and r is the radius.
Given:
Mass of the merry-go-round (m_merry) = 300 kg
Radius of the merry-go-round (r_merry) = 2.4 m
Using the formula, we can calculate the moment of inertia (I_merry) of the merry-go-round:
I_merry = 0.5 * m_merry * r_merry^2
I_merry = 0.5 * 300 kg * (2.4 m)^2
I_merry = 864 kg*m^2
Next, we need to convert the given angular velocity (ω_merry) of the merry-go-round from rpm to radians per second. Since 1 revolution is equal to 2π radians, we can use the conversion factor: 1 rpm = 2π/60 rad/s.
Given:
Angular velocity of the merry-go-round (ω_merry) = 22 rpm
Using the conversion factor, we can calculate ω_merry in rad/s:
ω_merry = 22 rpm * 2π/60 rad/s
ω_merry = 11π/15 rad/s
Now, let's find John's initial momentum just before he jumps. The momentum of an object can be calculated using the formula: momentum = mass * velocity.
Given:
Mass of John (m_john) = 25 kg
Velocity of John (v_john) = 5.0 m/s
Momentum of John (p_john) = m_john * v_john
p_john = 25 kg * 5.0 m/s
p_john = 125 kg*m/s
The principle of conservation of angular momentum states that the total angular momentum of a system before an event is equal to the total angular momentum after the event, provided no external torques are acting on the system.
Therefore, we can write the equation:
I_merry * ω_merry_before_jump = (I_merry + I_john) * ω_after_jump
Where:
I_john = 0.5 * m_john * r^2 (moment of inertia of John)
ω_after_jump = angular velocity of the merry-go-round after John jumps on
Now, let's solve for ω_after_jump:
I_merry * ω_merry_before_jump = (I_merry + I_john) * ω_after_jump
I_merry * ω_merry_before_jump = (I_merry + 0.5 * m_john * r^2) * ω_after_jump
ω_after_jump = (I_merry * ω_merry_before_jump) / (I_merry + 0.5 * m_john * r^2)
Substituting the known values:
ω_after_jump = (864 kg*m^2 * (11π/15 rad/s)) / (864 kg*m^2 + 0.5 * 25 kg * (2.4 m)^2)
Now we can solve for ω_after_jump by calculating the numerator and denominator separately:
Numerator = 864 kg*m^2 * (11π/15 rad/s)
Denominator = 864 kg*m^2 + 0.5 * 25 kg * (2.4 m)^2
After performing the calculations, we find that the numerator is equal to 101.864π rad/s and the denominator is equal to 1952.4 kg*m^2.
Therefore, ω_after_jump ≈ (101.864π rad/s) / (1952.4 kg*m^2)
Simplifying, ω_after_jump ≈ 0.0519π rad/s
Finally, let's convert ω_after_jump from rad/s to rpm:
1 rpm ≈ 0.1592π rad/s
Therefore, ω_after_jump in rpm ≈ 0.0519π rad/s * (1 rpm / 0.1592π rad/s)
Simplifying, ω_after_jump ≈ 0.325 rpm
So, the merry-go-round's angular velocity, after John jumps on, is approximately 0.325 rpm.
The angular momentum of merry-go-round PLUS John remains the same before and after he jumps on.
The moment of inertia of the merry-go-round is
I = (1/2) M R^2 = 216 kg*m^2.
The initial angular velocity of the merry-go-round is
w1 = 22*2*pi/60 = 2.30 rad/s
The angular momentum conservation equation is:
I*w1 + m*R*v = (I + mR^2)*w2
where m is John's mass.
498 + 150 = (216 + 36)*w2
w2 = 2.57 rad/s
So the Merry-Go-Round speeds up.
Finally, convert the w2 value to rpm.