A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 12 in 9.0 . Assume the merry-go-round is a uniform disk of radius 2.2 and has a mass of 650 , and two children (each with a mass of 29 ) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque.

Question

A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 12 in 9.0 . Assume the merry-go-round is a uniform disk of radius 2.2 and has a mass of 650 , and two children (each with a mass of 29 ) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque.

Answer 1

Step One: Convert rpm to radians per second

(12 rpm)(2pi rad/1rev)(1min/60s) = 1.26 rad/s = w

Step Two: Determine the Angular Acceleration

w = alpha t + w

1.26 rad/s = alpha (9.0 s) + 0

Alpha = 0.140

Step Three: Determine the moment of inertia of the disk and children

Moment of inertia = Inertia of disk + inertia of child + inertia of child = 1/2mass of disk * r^2 + 2mass of child * r^2.

So:
I = 1/2(650 kg)(2.2)^2 +2(29kg)(2.2)^2

Inertia total = 1573kg + 280.72kg = 1853.72 kg

Step Four: Determine the magnitude of the torque.

torque = Inertia * alpha

torque = (0.140)(1853.72) = 259.5208

So rounded, torque = 260 N.

Some questions also required the following: What force is required at the edge?

If so, force can be found using the following equation:

torque = rFsin(theta)

260N = (2.2)Fsin(90)

Sin(90) = 1

Therefore

F = 260N/2.2

F = 118N

Hope that helps!

Answer 2

Well, it seems like you want to calculate the torque required to accelerate the merry-go-round. Before I do that, let me explain the situation a bit.

So, it looks like this teenager is having a fun time pushing the merry-go-round. But in order to calculate the torque required, we need a little bit of physics.

To calculate the torque, we need to know the moment of inertia of the merry-go-round. The moment of inertia of a uniform disk is given by the formula I = (1/2) * m * r^2, where m is the mass and r is the radius.

So, let's calculate the moment of inertia of the merry-go-round:
I = (1/2) * 650 * (2.2)^2

Now, we also need to know the angular acceleration of the merry-go-round. The formula for angular acceleration is given by the formula a = alpha * r, where a is the tangential acceleration, alpha is the angular acceleration, and r is the radius.

So, the angular acceleration can be calculated as:
a = (2 * pi * f)^2 * r

Now, we can finally calculate the torque using the formula:
torque = I * alpha

So, putting it all together, the torque required to produce the acceleration is:
torque = [(1/2) * 650 * (2.2)^2] * [(2 * pi * 12)^2 * 2.2]

I hope that answers your question! And remember, physics can be merry-go- sometimes, but it's always spinning with knowledge!

Answer 3

To calculate the torque required to produce the acceleration, we can use the equation:

Torque (τ) = Moment of inertia (I) * Angular acceleration (α)

We can first calculate the moment of inertia of the merry-go-round, assuming it is a uniform disk. The moment of inertia for a uniform disk is given by the equation:

I = (1/2) * m * r^2

Where:
m = mass of the merry-go-round
r = radius of the merry-go-round

Given information:
merry-go-round mass (m) = 650 kg
merry-go-round radius (r) = 2.2 m

Substituting the values into the equation, we get:

I = (1/2) * 650 kg * (2.2 m)^2

Now we can calculate the moment of inertia (I).

I = (1/2) * 650 kg * (2.2 m)^2
= 786.5 kg * m^2

Next, we need to calculate the angular acceleration (α). The angular acceleration is related to the frequency by the equation:

ω = 2πf

Where:
ω = angular velocity (in radians per second)
f = frequency (in Hz)

Given information:
Frequency (f) = 12 Hz

Substituting the value into the equation, we get:

ω = 2π * 12 Hz
= 24π rad/s

Now we can calculate the angular acceleration (α). The angular acceleration is the rate of change of angular velocity and is given by the equation:

α = ω / Δt

Where:
α = angular acceleration (in rad/s^2)
ω = angular velocity (in rad/s)
Δt = time interval (in seconds)

Given information:
Time interval (Δt) = 9.0 s

Substituting the values into the equation, we get:

α = (24π rad/s) / (9.0 s)

Now we can calculate the angular acceleration (α).

α = (24π rad/s) / (9.0 s)
≈ 8.49 rad/s^2

Finally, we can calculate the torque (τ) required using the equation:

τ = I * α

Substituting the values, we get:

τ = (786.5 kg * m^2) * (8.49 rad/s^2)

Now we can calculate the torque (τ) required.

τ ≈ 6670 N*m

Answer 4

To calculate the torque required to produce the acceleration of the merry-go-round, we need to use the equation for torque:

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 of the merry-go-round. We can use the formula for the moment of inertia of a uniform disk:

I = (1/2) * m * r^2

Where m is the mass of the disk and r is the radius of the disk.

In this case, the mass of the disk is given as 650 kg and the radius is given as 2.2 m. Plugging these values into the formula, we get:

I = (1/2) * 650 kg * (2.2 m)^2
= 800.3 kg·m²

Next, we need to calculate the angular acceleration. We can use the formula:

α = (Δω) / Δt

Where Δω is the change in angular velocity and Δt is the time taken for the change.

In this case, the initial angular velocity is given as 0 (since the merry-go-round starts from rest) and the final angular velocity is given as 12 rad/s. The time taken for the change is given as 9.0 s. Plugging these values into the formula, we get:

α = (12 rad/s - 0) / 9.0 s
= 1.33 rad/s²

Finally, we can calculate the torque using the equation:

τ = I * α

Plugging in the values we calculated, we get:

τ = 800.3 kg·m² * 1.33 rad/s²
= 1,064.1 N·m

So, the torque required to produce the acceleration of the merry-go-round is 1,064.1 N·m.