Sunday

April 20, 2014

April 20, 2014

Posted by **cara** on Tuesday, April 19, 2011 at 4:32pm.

- Physics -
**Sara**, Tuesday, March 6, 2012 at 4:30pmStep 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!

- Physics -
**Sara**, Tuesday, March 6, 2012 at 4:32pmStep 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!

- Physics -
**Sara**, Tuesday, March 6, 2012 at 4:38pmStep 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!

**Related Questions**

physics - A dad pushes tangentially on a small hand-driven merry-go-round and is...

Physics - A horizontal 873 N merry-go-round of radius 1.11 m is started from ...

Physics - A merry-go-round with r = 4m and a perfect frictionless bearing is ...

physics - You wish to accelerate a small merry-go-round from rest to a ...

Physics - A merry-go-round in the park has a radius of 1.8 m and a rotational ...

physics - A merry-go-round in a playground has a mass of 200 kg and radius of 1....

physics - A horizontal 810-N merry-go-round of radius 1.70 m is started from ...

physics - A horizontal 790-N merry-go-round of radius 1.30 m is started from ...

physics - A merry-go-round in the park has a radius of 1.8 m and a rotational ...

physics - A merry go round in the park has a radius of 1.8 m and a rotational ...