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**11 Rotational Dynamics and Static Equilibrium**

Torque: τ τ=rF┴ SI unit: N*m

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F F┴ F// θ τ=rF┴=r(Fsinθ) θ : the angle between force r and F. τ direction measures by right hand rule, the four fingers are curled from the direction of r toward the direction of F, then the thumb points in the direction of the torque.

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**τ>0, counter clockwise**

F θ τ>0, counter clockwise τ<0, clockwise r F θ Ex. F1=10N 300 400 F2=8N r r1=0.2m r2=0.1 m τ 1, τ2

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**Torque: τ angular acceleration α Force: F acceleration a s=rθ F┴=mat=mrα r F┴=mr2α τ=Iα **

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**α: angular acceleration**

Newton’s 2nd law for rotational motion : sum of torque I: moment of inertia α: angular acceleration

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M=0.1 kg r=0.1 m m=0.2kg Find α ?

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**Zero Torque and static equilibrium**

W1 W2 N

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Ex. A 5.00-m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board, as shown below; the other is 1.50 m away. Find the forces exerted by the pillars when a 90.0-kg diver stands at the far end of the board.

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A hiker who has broken his forearm rigs a temporary sling using a cord stretching from his shoulder to his band. The cord holds the forearm level and makes an angle of with the horizontal where it attaches to the hand. Considering the forearm and hand to be uniform, with a total mass of 1.31-kg and a length of 0.3 m, find (a) the tension in the cord and (b) the horizontal and vertical components of the force, f, exerted by the humerus (the bone of upper arm) on the radius and ulna (the bones of the forearm).

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Ladder An 85 kg person stands on a lightweight ladder, as shown. The floor is rough, hence, it exerts both a normal force, f1, and a frictional force, f2, on the ladder, The wall, on the other hand, is frictionless, it exerts on a normal force, f3, find the magnitudes of f1, f2, and f3.

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**11-4 Center of mass and balance**

m1gx1-m2gx2=0

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**11-5 Dynamic Application of torque**

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**A 0. 31-kg cart on a horizontal air track is attached to string**

A 0.31-kg cart on a horizontal air track is attached to string. The string passes over a disk-shaped pulley of mass 0.08kg, and radius 0.012m and is pulled vertically downward with a constant force of 1.1N Find (a) the tension in the string between the pully and the cart and (b) the acceleration of the cart.

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11-6 Angular momentum Recall: F=ma=mΔv/Δt=Δp/Δt F=0, p=constant τ= Iα=IΔω/Δt=ΔL/Δt L=Iω : angular momentum

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**Find the angular momentum of (a) a 0**

Find the angular momentum of (a) a 0.13-kg Frisbee (considered to be a uniform disk of radius 7.5 cm) spinning with an angular speed of 1.15 rad/s and (b) a 95-kg person running with a speed of 5.1 m/s on a circular track of radius 25m.

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**11-7 Conservation of Angular Momentum**

τ=0 ΔL=0 Li=Lf

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**Li=Lf Iiωi=Ifωf ωf=ωiIi/If**

Ii>If so ωf>ωi

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The student holds his arms outstretched and spins about the axis of the stool with an angular speed of 10 rad/s. The moment of inertia in this case is 6.00 kgm2, while still spinning, the student pulls his arms in to his chest, reducing the moment of inertia to 2 kgm2, what is the student’s angular speed now?

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**A star of radius R=2. 3x108 m rotates with an angular speed ω=2**

A star of radius R=2.3x108 m rotates with an angular speed ω=2.4x10-6 rad/s, It this star collapses to a radius of 20.0km, find its final angular speed. (Treat the star as if it were a uniform sphere I=2/5MR2, and assume that no mass is lost as the star collapse)

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The turntable with a moment of inertia It is rotating freely with an initial angular speed ω0. A record, with a moment of inertia Ir and initially at rest, is dropped straight down onto the rotating turntable, as in Figure.

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**11-8 Rotational work and power**

W=FΔx Δx=RΔθ W=FΔx=FRΔθ τ=RF Work Done by Torque W=τΔθ Power Produced by a Torque: P=W/Δt=τΔθ/Δt=τω

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**It takes a good deal of effort to make homemade ice cream**

It takes a good deal of effort to make homemade ice cream. (a) If the torque required to turn the handle on an ice cream maker is 5.7 Nm, how much work is expended on each complete revolution of the handle? (b) How much power is required to turn the handle if each revolution is completed in 1.5s?

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