{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework6-F08 - Homework 6 Chapt 8 9 Accelerating along a...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework 6 Chapt 8 & 9 Accelerating along a Racetrack Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity v harpoonrightnosp is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector v harpoonrightnosp at each point along the curved path and is directed toward the center of curvature of the path. The magnitude of the acceleration of an object that moves with constant speed v along a circular path of radius r is given by . 8.35. Model: Use the particle model for the ball in circular motion. Visualize: Solve: (a) The mass moves in a horizontal circle of radius 20 cm. r = The acceleration a arrowrightnosp and the net force vector point to the center of the circle, not along the string. The only two forces are the string tension , T arrowrightnosp which does point along the string, and the gravitational force G . F arrowrightnosp These are shown in the free-body diagram. Newton’s second law for circular motion is G cos cos 0 N z F T F T mg θ θ = - = - = 2 sin r r mv F T ma r θ = = = From the z -equation, ( ) ( ) 2 0.500 kg 9.8 m/s 5.00 N cos cos11.54 mg T θ = = = ° (b) We can find the tangential speed from the r -equation: sin 0.63 m/s rT v m θ = = The angular speed is 0.63 ms 3.15 rad/s 30 rpm 0.20 m v r ω = = = = 8.46.
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}