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Inertia of door vs. braking torque?

  1. Jul 10, 2014 #1
    Hi. I'm designing what effectively is a large door on hinges. The door has the following properties

    Mass = 336 kg
    Width = 1.8m

    The door rotates from closed position through 45deg to open position. In order to 'control' or damp the rotation of the door, I'm fitting a 3Nm friction clutch on the hinge axis.

    What I'm struggling to calculate is - Is the friction in the friction clutch high enough to effectively control the rotation of the door without it slamming into it's endstops in the open and closed positions?

    I know the torque. I can calculate the door's moment of inertia. I can guess at the force that an operator might push the door - 10N? If 10N is reasonable, then the torque exerted is 18Nm. So, the friction clutch will easily be exceeded, which is good because otherwise the operator wouldn't be able to open/close the door.

    But where I'm struggling is to understand how the friction will retard the rotation of the door to prevent it slamming into the endstops.

    Any help much appreciated. Thanks.
  2. jcsd
  3. Jul 10, 2014 #2


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    This is a rather heavy door, but you have not stated how it is mounted on its hinges. Is this a normal door with the hinges in-line vertically? Or is it mounted in some other manner?

    In any event, 10N doesn't seem like enough force to cause this door to slam even without any friction clutches mounted. If you want to slam a room door, for example, you don't give it a gentle 10 N nudge, you grab the door with your hand and swing your arm across your body.
  4. Jul 10, 2014 #3
    It's essentially a rack of equipment, but to all intents and purposes it's a heavy door.

    There are 2 hinges in line vertically.

    My concern is that due to the large inertia, once the door starts moving it won't easily stop.
  5. Jul 10, 2014 #4


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    Will adding a friction clutch make the door harder to open? You might just encourage the operator to pull harder.

    If you know the torque that can be applied and the moment of inertia you can work out how fast an operator could get the door rotating from a standing start.

    Torque = moment of inertia * angular acceleration

    Perhaps assume he pulls hard and keeps accelerating the door until it reaches say 45 degrees open. Work out how fast it's rotating at that point

    Then perhaps assume that at the 45 degree point he stops trying to accelerate the door and allows it to coast towards the end stops. It will decelerate due to the torque from the friction clutch and you can work out how fast it will be going when it hits the end stops (or even if it will hit the end stop).

    I'm thinking you might do better to build a shock absorber and damper into the end stops because the forces involved will also depend on the stopping distance.
  6. Jul 10, 2014 #5


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    For this purpose the resistance should be velocity dependent, rather than constant:

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