Circular Motion and Gravitation3

# Circular Motion and Gravitation3 - Circular Motion and...

This preview shows pages 1–3. Sign up to view the full content.

Circular Motion and Gravitation home - about - terms - credits - feedback Physics Tutorial Minds on Physics Multimedia Physics Studios Shockwave Physics Studios The Review Session 1-D Kinematics Newton's Laws of Motion Vectors and Projectiles Forces in Two Dimensions Momentum and Collisions Work, Energy and Power Circular and Satellite Motion Static Electricity Electric Circuits Waves Sound and Music Light and Color Reflection and Mirrors Refraction and Lenses Physics Help Curriculum Corner The Laboratory » The Physics Classroom » Review Session » Circular Motion and Gravitation Circular Motion and Gravitation Review Navigate to Answers for: Questions #1-#14 Questions #15-#28 Questions #29-#40 [ #29 | #30 | #31 | #32 | #33 | #34 | #35 | #36 | #37 | #38 | #39 | #40 ] Part E: Problem-Solving 29. A roller coaster car loaded with passengers, has a mass of 500 kg; the radius of curvature of the track at the bottom point of a dip is 12 m. The vehicle has a speed of 18 m/s at this point. a. In the space below, draw a free-body diagram for the car (label forces according to type). 2. Calculate the acceleration and the net force acting upon the car. PSYW 3. Calculate the force exerted on the vehicle by the track? PSYW Answers: The free-body diagram (part a) is shown at the right. The acceleration of the car can be computed as follows: a = v 2 /R = (18.0 m/s) 2 /(12.0 m) = 27.0 m/s 2 (part b) The net force can be found in the usual manner: F net = m a = (500. kg) (27.0 m/s 2 ) = 13500 N (part b) Since the center of the circle (see diagram) is above the riders, then both the net force and the acceleration vectors have an upward direction. The force of http://www.physicsclassroom.com/reviews/circles/cpmans3.cfm (1 of 10) [4/12/2552 20:37:25]

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

View Full Document
Circular Motion and Gravitation gravity is downwards, so the net force is equal to the upward force minus the downward force: F net = F norm - F grav where F grav = m g = (500. kg) (9.8 m/s/s) = 4900 N Thus, F norm = F net + F grav = 13500 N + 4900 N = 18400 N (part c) Useful Web Links Roller Coasters and Amusement Park Physics [ #29 | #30 | #31 | #32 | #33 | #34 | #35 | #36 | #37 | #38 | #39 | #40 ] 30. What is the acceleration of a piece of dust on an old-fashioned record album, 15 cm from the center, if the record is spinning at 33.3 rpm? Answer: 1.82 m/s/s To find acceleration, the speed and the radius must be known. The radius is given; the speed can be computed as distance per time. The dust moves a distance equivalent to 33.3 circumferences in 60 seconds. So v = 33.3 2 pi (0.15 m) / (60 s) = 0.523 m/s. The acceleration can now be computed using the centripetal acceleration equation: a = v 2 /R = (0.523 m/s) 2 / (0.15 m) = 1.82 m/s/s Useful Web Links Mathematics of Circular Motion [ #29 | #30 | #31 | #32 | #33 | #34 | #35 | #36 | #37 | #38 | #39 | #40 ] 31. What is the tension in a 0.500 meter rope which carries a 2.50 kg bucket of water in a vertical circle with a
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

Circular Motion and Gravitation3 - Circular Motion and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online