A playground merry-go-round made in the shape of a solid disk, has a diameter of 2.63 m and a mass of 353.9 kg. Two children, each of mass 30.5 kg, sit on opposite sides at the edge of the platform. Approximate the children as point masses.
(a) What torque is required to bring the merry-go-round from rest to 26 rpm in 19.7 s?
(b) If two other bigger children are going to push on the merry-go-round rim to produce this acceleration, with what force magnitude must each child push?
From rotational kinematics (n=26rpm=26/60 rev/s)
2•π•n = ε•t
ε =2•π•n/t =2•π•26/60•19.7 =0.138 rad/s^2
Moment of inertia (platform + 2 point masses):
I = M•R^2/2 + 2•M•R^2 = (1.315)^2•(353.9/2 + 2•30.5) = 412 kg•m^2
Newton’s 2 Law for rotation
M=I• ε =412•0.138 = 56.8 N•m.
The torque of coupled forces is
M =F•r, where
r is the distance between the points where the forces are applied (here r =D)
F = M/D =56.8/2.63 = 149 N
To solve this problem, we will use the following formulas:
(a) The torque required to bring the merry-go-round from rest to a certain angular velocity can be calculated using the formula:
Torque = Moment of Inertia * Angular acceleration
(b) The force required to produce this acceleration at the rim of the merry-go-round can be calculated using the formula:
Force = Mass * Acceleration
Let's calculate:
(a) First, we need to find the moment of inertia of the merry-go-round. The moment of inertia of a solid disk is given by the formula:
Moment of Inertia = (1/2) * Mass * Radius^2
where Mass is the mass of the merry-go-round and Radius is its half the diameter.
Mass = 353.9 kg
Radius = 2.63 m / 2 = 1.315 m
Moment of Inertia = (1/2) * 353.9 kg * (1.315 m)^2
Now, we need to convert the angular velocity from rpm to radians per second:
Angular velocity = 26 rpm * (2π radians / 1 minute)
= 26 * 2π / 60 radians per second
Next, we need to find the angular acceleration, which can be calculated using the formula:
Angular acceleration = (Final angular velocity - Initial angular velocity) / Time
Final angular velocity = 26 * 2π / 60 radians per second
Initial angular velocity = 0 (since the merry-go-round starts from rest)
Time = 19.7 s
Now, we can calculate the torque:
Torque = Moment of Inertia * Angular acceleration
(b) To find the force required to produce this acceleration, we will use the formula:
Force = Mass * Acceleration
where Mass is the mass of the two children and Acceleration is the angular acceleration calculated in part (a).
Mass = 30.5 kg * 2 (since there are two children)
Acceleration = Angular acceleration
Now, we can calculate the force magnitude required for each child:
Force = Mass * Acceleration
Let's calculate:
(a) Moment of Inertia:
Moment of Inertia = (1/2) * 353.9 kg * (1.315 m)^2
Angular velocity:
Angular velocity = 26 * 2π / 60 radians per second
Angular acceleration:
Angular acceleration = (26 * 2π / 60 radians per second - 0 radians per second) / 19.7 s
Torque:
Torque = Moment of Inertia * Angular acceleration
(b) Force:
Force = (30.5 kg * 2) * Angular acceleration
Now you can substitute the values and calculate the torque and force.
To answer these questions, we need to use the formulas for torque and rotational motion.
(a) To calculate the torque, we need to find the moment of inertia of the merry-go-round and then use the equation:
Torque = Moment of Inertia * Angular Acceleration
1. Moment of Inertia (I):
The moment of inertia depends on the shape and mass distribution of the object. For a solid disk rotating on its axis, the moment of inertia is given by the equation:
I = (1/2) * m * r^2
where:
m is the mass of the merry-go-round (353.9 kg),
r is the radius of the merry-go-round (half of the diameter = 2.63 m / 2 = 1.315 m).
So, I = (1/2) * 353.9 kg * (1.315 m)^2
2. Angular Acceleration (α):
The angular acceleration is given by the equation:
α = Δω / Δt
where:
Δω is the change in angular velocity (from rest to 26 rpm = (26 revolutions/minute) * (2π radians/revolution) / (60 seconds/minute)),
Δt is the change in time (19.7 seconds).
Calculate the angular acceleration (α) using the above equation.
Finally, multiply the moment of inertia (I) with the angular acceleration (α) to find the torque (τ).
(b) To find the force with which the children must push, we will use Newton's second law for rotational motion:
τ = I * α = (F * r) * α
where:
τ is the torque,
r is the radius of the merry-go-round,
α is the angular acceleration, and
F is the force acting at the rim.
As two bigger children are pushing on the rim of the merry-go-round, each child exerts a force F. So, we need to find the force (F).
Divide both sides of the equation by r and rearrange the equation to solve for F:
F = (τ / (r * α))
Substitute the given values for torque (τ), radius (r), and angular acceleration (α) into the equation to calculate the force magnitude (F) required for each child.