webct2 - disk 1.A red spot is painted on the edge of a disk...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
disk 1.A red spot is painted on the edge of a disk of radius 30.0 cm. The spot has a constant speed of 15.0 m/s. How many turns does the red spot describe in one second? 15 m/s 7.96 turns/s 2 2 2 (0.30 m) v f R ω ππ π == = = 2. What is the angular velocity of the spot, in rad/s? 15 m/s 50 rad/s 0.30 m v R = 3. What is the centripetal acceleration of the spot, in m/s 2 ? a c = r 2 = 750 rad/s 2 4. What is the tangental acceleration of the spot, in m/s 2 ? Since the speed is constant, the tangential acceletation is zero. 5. A yellow spot is painted on the same disk at 15 cm from the center. What is the linear speed of the yellow spot, in m/s? () ( ) 0.15 m 50 rad/s 7.5 m/s vR = Since circumference = 2 π R , the yellow spot covers half the distance covered by the red spot in the same amount of time. Thus, red yellow 2 v = v . 6. What is the angular velocity of the yellow spot, in rad/s? Angular velocity is independent of radius (every point on the disk does the same number of turns per second), so 50 rad/s again.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
7. How long does it take for the red spot to make one turn, in s? 12 0.126 s T f π ω == =
Background image of page 2
a not ct 1. The time-dependent acceleration of an object is a = 3 t i m/s 2 . At t = 0, the velocity is (0,-2) m/s and the object is at ( x,y ) = (0,4) m . Find is position at t = 3 s. Since we are given the acceleration, two integrations will be required to get to the position. Note that the acceleration is NOT constant, so the equations for motion at constant acceleration cannot be used. () 2 0 00 33 2 0 3 ˆˆ ˆˆ 23 2 2 3 ˆˆ ˆ ˆ ˆ ˆ ˆ 42 4 2 4 2 22 2 tt v v adt j tidt j t i r r v d tj jt i d t ji i t j =+ = −+ =  =+−+ =−+ = +−   ∫∫ !! ! ! ˆ At t = 3 s, 3 3 ˆ (3 s) 4 2(3) 4 2 m 2 ri j i =− ! ˆ j
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2. Which of the following options is the best representation of the trajectory of the object? At t=0, the object is moving down, so it must be either B or E. In E, the object reaches a certain (negative) value of y and then turns around, which would imply an acceleration in the y direction. So it’s not E. In B, instead, we have an increasing velocity in the +x direction, as expected from the acceleration.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

webct2 - disk 1.A red spot is painted on the edge of a disk...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online