5. acceleration – change in speed over time (vector quantity) TWO types; a. Linear acceleration – speed up or slow down b. Centripetal acceleration – change direction B. Centripetal acceleration (ac) – acceleration changes due to change in direction. 1. Centripetal means center seeking 2. ac is always directed toward the center of the curved path (circle) 3. If an object is moving in a circle it will always have a centripetal acceleration 4. ac
Free Force Kinematics Classical mechanics
to Review or Learn: Memorize/use/convert SI units Apply Newton’s laws to circular motion Dispel myths about circular motion I am responsible for Vocabulary to know Centripetal Tangential Gravitational Torque I am responsible for Vocabulary to know Centripetal Tangential Gravitational Torque 7.1 ROTATIONAL MOTION A car is driving at a constant speed of 13.6 m/s (about 30 mph)‚ moving counter-clockwise on a circular track as shown
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Background Information The Giant Drop The Giant Drop is a vertical free fall‚ looming 119m meters above the ground. Carried by a mechanical lift to the very top‚ it then plummets‚ reaching up to 135km/hr due to the acceleration of gravity‚ before finally coming to a stop with the magnetic braking system (Burton‚ 2009). A rider on this type of design will experience three phases of apparent weight: the lifting‚ falling and braking stages. At first‚ the rider will feel
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1 1 2 3 C Motion I 7 (a) From 1 January 2009 to 10 January 2009‚ the watch runs slower than the actual time by 9 minutes. Therefore‚ when the actual time is 2:00 pm on 10 January 2009‚ the time shown on the watch should be 1:51 pm on 10 January 2009. Practice 1.1 (p. 6) D (a) Possible percentage error 10 −6 = × 100% 24 × 3600 = 1.16 × 10 % 1 (b) = 1 000 000 days 10 −6 –9 It would take 1 000 000 days to be in error by 1 s. (b) Percentage error 9 = × 100% 9 × 24 × 60 = 6.94 × 10–2%
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Processing Raw Data: Conclusion: It is hypothesized that the tension found in the string based on the mass of the washers would be equal to the tension derived from the formula for centripetal force‚ because it is the same string and therefore the same tension. Our findings supported this as the centripetal force formula gave us a tension that is very close to the tension found through the mass of the washers when you account for errors in measuring. The forces must have been equal‚ or the washers
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concurrently on a point. As the angle between these forces increases from 0° to 90°‚ the magnitude of their resultant (1) decreases (2) increases (3) remains the same 2. A car increases its speed from 9.6 m/s to 11.2 m/s in 4.0 s. The average acceleration of the car during this 4.0-second interval is (1) 0.40 m/s2 (3) 2.8 m/s2 (2) 2.4 m/s2 (4) 5.2 m/s2 3. What is the speed of a 2.5-kilogram mass after it has fallen freely from rest through a distance of 12 m? (1) 4.8 m/s (3) 30. m/s
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a circle affects how much centripetal force affects an object. Centripetal force plays a major part in every day life. This force is the reason for the sensation of being ‘pushed’ into the car door that passengers in a car feel when the driver makes a sharp left turn (The Centripetal Force Requirement‚ n.d.). It is the reason people feel almost weightless on rollercoasters‚ and on an even larger scale‚ it is what keeps the moon orbiting around Earth. Without centripetal force‚ cars would not be able
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particle starts from the origin at t = 0 with a velocity of 6.0[pic] m/s and moves in the xy plane with a constant acceleration of (-2.0[pic] + 4.0[pic]) m/s2. At the instant the particle achieves its maximum positive x coordinate‚ how far is it from the origin? [pic] 2 At t = 0‚ a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. Its acceleration is given by [pic] = (3.0[pic] - 2.0[pic]) m/s2. At the instant the particle reaches its maximum y coordinate how far
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Chapter 4 Problems 1‚ 2‚ 3 = straightforward‚ intermediate‚ challenging Section 4.1 The Position‚ Velocity‚ and Acceleration Vectors 1. A motorist drives south at 20.0 m/s for 3.00 min‚ then turns west and travels at 25.0 m/s for 2.00 min‚ and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip‚ find (a) the total vector displacement‚ (b) the average speed‚ and (c) the average velocity. Let the positive x axis point east. 2. A golf ball is hit off a tee at the
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very long time [d] the velocity of the particle will become u/2 after time 1/α Q.2 A particle moves along the xaxis as x = u(t-1)2 + a(t-3)3 [a] initial velocity of the particle is u [b] the acceleration of the particle is a [c] the acceleration of the particle is 2a [d] the particle is at the origin at time t=3 seconds Q.3 A particle is projected vertically upwards [a] the speed decreases uniformly with distance [b] the speed decreases
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