Please show me the work so I understand 1). A meter stick is rotated about the e

Question

Please show me the work so I understand

1). A meter stick is rotated about the end labeled 0.00 cm, so that the other end of the stick moves through an arc length of 8.80 cm. Through what approximate arc length does the 25.0 cm mark on the stick move? D), 25 cm A). 2.2 cm B). 4.4 cm C). 8.8 cm 2). You are in a tall building located near the equator. As you ride an elevator from the ground floor to the top floor, your tangential speed due to the earth's rotation C). increases B). decreases A). does not change D). increases when the speed of the elevator increases and decreases when the speed of the elevator decreases 30. A Ferris wheel of radius 25 m is spinning at a constant angular speed of 12 s The linear speed of a point on the circumference of the wheel is D). 180 m/s B). 5.2 m/s C). 0.5 m/s A). 300 m/s 4). At an instant when the magnitude of the linear acceleration of a point on the circumference of the above mentioned Ferris wheel is increasing at a rate of 0.5 m/s, the magnitude of the angular acceleration of the wheel is D). 0.4 rad s C). 12.5 rad s A). 0.125 rad s B). 0.02 rad s 5). The disk of radius b and mass m shown at right is free to rotate in a vertical plane without friction about a fixed horizontal axel through point O. Three forces. Fi, F2, and F3, are applied to the disk as shown. The net torque on the disk about the axis of rotation is B). Fla cos 30 F2b Fab A). Fia F2b F3b D). Fia cos30 F2b F3b C). Fia Fab F3b

Explanation / Answer

1. angle through which stick moves = arc/length = 0.088m/1m = 0.088 rad

therefore, at .25m arc length = angle*length = 0.088*0.25 = 0.022m or 2.2cm (A)

2. as we go higher the effective radius increases, since angular speed is same and  

v=?.r tangential speed increases to keep  ? constant (C)

3. angular velocity in radians ? = 12*3.14/180 = 0.21 rad/sec

velocity at circumference = r. ? = 25*0.21 = 5.25 m/s (B)

4. angular acc = linear acc/radius = 0.5/25=0.02 rad/s2 (B)

5. torque due to F1 = F1.a anticlockwise

torque due to F2 = F2.b clockwise

      torque due to F3 = F3.b clockwise

   therefore total torque = F1.a-F2.b-F3.b (A)

6. Torque due to 25N force = 0 as it is passing through axis of rotation

      Torque due to 30N force = 30*3*sin45 = 63.64 Nm anticlockwise

  Torque due to 10N force = 10*3*sin25 = 12.67 Nm clockwise
Therefore total torque = 63.64-12.67 = 50.97 Nm (A)

7. torque = position vector X force = (2i+3j)X(3i-2k) = -6i + 4j -9k (B)

8. perpendicular distance of force A about O is 2 units

perpendicular distance of force B about O is 3 units

  perpendicular distance of force C about O is 2 units

Therefore  A= ?c< ?B (B)

9. Taking moment of torque about point D U.L3 = Fg.(L2+L3)

U.1 = 3*9.8*2.5 U = 73.5N (D)

10.    balancing forces along vertical U = D + Fg     

D = 73.5-39.8 = 44.1N (A)

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