1 / 15

Torque

Torque. AP Physics 1. Causing Rotational Motion. In order to make an object start rotating about an axis, a torque is required. Torque includes not only the amount of force applied but also the distance from the axis to which the force is applied. Distance from axis is called the lever arm.

cdick
Download Presentation

Torque

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Torque AP Physics 1

  2. Causing Rotational Motion • In order to make an object start rotating about an axis, a torque is required. • Torque includes not only the amount of force applied but also the distance from the axis to which the force is applied. • Distance from axis is called the lever arm. • In some cases, the lever arm is the distance from the center of mass. • For symmetrical objects the center of mass is in the center.

  3. Torque Definition • The combination of amount of force and distance from axis is called Torque (τ). • A torque is necessary in order to provide angular acceleration. • τ = FrsinѲ • Units = N-m • Direction = • Positive torque is clockwise (CW) • Negative torque counter-clockwise (CCW)

  4. Torque Equation • In order to use this equation, the force must be perpendicular to the axis of rotation. • τ = FrsinѲ

  5. Example 1

  6. Example 2

  7. Example 3

  8. Rotational Equilibrium • Occurs when total torques acting balance and no rotational motion occurs. • In order for equilibrium to exist, the total torque acting in the clockwise direction must balance the total torque acting in the counterclockwise direction.

  9. See-Saw • When stationary, the total τ clockwise is equal to the τ counterclockwise • If kid A pushes up with more force, the τ is no longer balanced and the see-saw begins to rotate clockwise with an angular velocity.

  10. Example 4

  11. Example 4 • Στ = CCW – CW = 0 • Therefore, CCWτ = CWτ • CCWτ = Person A = Fr = (500)r • CWτ = Person B = Fr = (1000)(1) • Setting these equal and solving for r gives: 500r = 1000 r = 2 m

  12. Example 5

  13. Example 4 So the bar should be placed 2.25 m from the left or 3.75 m from the right. In this example, you have to take into account the weight of the board since the fulcrum is not under the center of mass. CCW τ = CW τ (900)(r) = (200)(3-r) + (500)(6-r) (900)(r) = 3600 – 700(r) 1600(r) = 3600 r = 2.25 m

  14. FT 30° 30° Eat at Joe’s Fcm = 100N Pivot point mg = 200N Example 6 A 20 kg sign hangs from a 2 meter long rod that has a mass of 10 kg and is supported by a cable at an angle of 30° as shown below. Determine the tension in the cable. (force at the pivot point is not shown because τ = 0)

  15. FT 30° Fbar = 100N Pivot point Fsign = 200N Example 6 ∑ τ = 0 τcable = τ bar + τ sign FT(2)sin30 =100(1) + (200)(2) FT = 500 N

More Related