Peoples Physics Concepts

http://www.ck12.org Chapter 9. Rotational Motion

Solution

To solve this problem, we will plug in the known values into Newton’s second law for rotation and solve forα.

When calculating the moment of inertia of the door, we’ll consider it a rod being rotated about one end. We’ll first

calculate the angular acceleration when pushing from the center of the door.

Στ=Iα

Fr=Iα

Fr=

1

3

mr^2 α

α=

3 Fr

mr^2

α=

3 ∗50 N∗.75 m

20 kg∗( 1 .5 m)^2

α= 2 .5 rad/s^2

Now we’ll calculate the angular acceleration when pushing from the end of the door.

α=

3 Fr

mr^2

α=

3 ∗50 N∗ 1 .5m

20 kg∗( 1 .5 m)^2

α=5 rad/s^2

Clearly, it is much faster and easier to open doors when pushing from the point farthest from the hinge.

Watch this Explanation

MEDIA

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Explore More

1. In the figure we have a horizontal beam of length,L, pivoted on one end and supporting 2000 N on the other.
   Find the tension in the supporting cable, which is at the same point at the weight and is at an angle of 30
   degrees to the vertical. Ignore the weight of the beam.


2. Two painters are on the fourth floor of a Victorian house on a scaffold, which weighs 400 N. The scaffold is
   3 .00 m long, supported by two ropes, each located 0.20 m from the end of the scaffold. The first painter of
   mass 75 kg is standing at the center; the second of mass, 65.0 kg, is standing 1.00 m from one end.