A child pushes tangentially on a small hand driven merry- go- round. She is able to accelerate it from rest to 18 rpm in 10 s. Modeling the merry go round as a uniform disk of radius 2.5 m and mass 780 kg, find the torque required to produce that acceleration. Neglect the frictional torque.
To find the torque required to produce the given acceleration, we can use the rotational analog of Newton's second law, which states that the torque applied to an object is equal to the moment of inertia multiplied by the angular acceleration.
The formula for the torque (τ) is given by:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
First, let's calculate the moment of inertia (I). For a uniform disk, the moment of inertia is given by the formula:
I = (1/2) * m * r^2
where m is the mass of the disk and r is the radius.
Given:
Mass (m) = 780 kg
Radius (r) = 2.5 m
Substituting the values into the formula, we get:
I = (1/2) * (780 kg) * (2.5 m)^2
Now, let's calculate the angular acceleration (α). Angular acceleration can be determined by converting the rotational speed to radians per second and dividing it by the time.
Given:
Rotational speed = 18 rpm
Time (t) = 10 s
To convert the rotational speed to radians per second, we need to multiply it by 2π (since 1 revolution = 2π radians).
Angular speed (ω) = (18 rpm) * 2π = 36π radians per minute
To convert from minutes to seconds, divide by 60:
Angular speed (ω) = (36π radians per minute) / 60 = (3/5)π radians per second
Finally, we can calculate the angular acceleration:
α = ω / t = [(3/5)π radians per second] / 10 s
By simplifying the equation, we get:
α = (3/50)π radians per second^2
Now, substitute the values of I and α into the torque formula:
τ = I * α = [(1/2) * (780 kg) * (2.5 m)^2] * [(3/50)π radians per second^2]
Simplifying the equation, we find:
τ = (1/2) * (780 kg) * (2.5 m)^2 * (3/50)π radians per second^2
Calculating this expression will give us the torque required to produce the given acceleration.
To find the torque required to produce the acceleration, we need to determine the moment of inertia of the merry-go-round.
The moment of inertia for a uniform disk rotating about its axis is given by the formula:
I = (1/2) * m * r^2
Where:
I = moment of inertia
m = mass of the disk
r = radius of the disk
Using the given values:
m = 780 kg
r = 2.5 m
I = (1/2) * 780 kg * (2.5 m)^2
I = 6093.75 kg * m^2
Next, we need to convert the final angular velocity from rpm to rad/s. Since 1 rpm is equal to (2π/60) rad/s:
ω = (18 rpm) * (2π/60) rad/s
ω = 3π/5 rad/s
The initial angular velocity is 0 since the merry-go-round starts from rest.
The angular acceleration (α) can be found using the formula:
α = Δω / Δt
Where:
Δω = change in angular velocity
Δt = change in time
Δω = (3π/5 rad/s) - 0 rad/s
Δω = 3π/5 rad/s
Δt = 10 s
α = (3π/5 rad/s) / (10 s)
α = 3π/50 rad/s^2
Now, we can use the equation τ = I * α to find the torque:
τ = (6093.75 kg * m^2) * (3π/50 rad/s^2)
τ = (18281.25π/50) N * m
Therefore, the torque required to produce the acceleration is approximately 364.63 N * m or 365 N * m (rounded to the nearest whole number).