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Physics 140 – Fall 2014 September 16 Projectile & Circular Motion wikimedia.org Reminder: • Mastering Physics assignment 1 due tonight at midnight 1 Questions on today’s youtube video? 3 Shooting a Monkey (or Bear) • There was a clever monkey who, seeing that a hunter was about to shoot, decided to drop from the tree at the instant that he saw the flash from the gun as the bullet left the barrel, and thus escape the bullet. The hunter aimed the gun directly at the monkey at the monkey and fired. What happened? Let’s find out! A. The bullet hit the monkey B. The bullet missed the monkey and passed above it C. The bullet missed the monkey and passed below it Shooting a monkey 1 2 gt 2 Tilted Rocket problem – part 1 (based on P3.47) A test rocket is launched by accelerating it along a 200.0-m incline at 1.25 m/s2 starting from rest at point A on the diagram. The incline rises at 35.0° above the horizontal, and at the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity (ignore air resistance). What is the speed of the rocket as it leaves the incline? A) B) C) D) E) 11.2 m/s 22.4 m/s 33.6 m/s 44.8 m/s 56.0 m/s v02 2as v0 2(1.25 m/s2 )(200. m) 22.4 m/s v0 6 Tilted Rocket problem – part 2 (based on P3.47) A test rocket is launched by accelerating it along a 200.0-m incline at 1.25 m/s2 starting from rest at point A on the diagram. The incline rises at 35.0° above the horizontal, and at the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity (ignore air resistance). v0 22.4 m/s What is the maximum height above the ground that the rocket Reaches? x0 = (200.0 m) cos(35.0°) = 163.8 m A) 67 m y0 = (200.0 m) sin(35.0°) = 114.7 m B) 94 m v0 C) 123 m D) 156 m E) 199 m (x0, y0) 7 At the top: H y0 v 0 v 2ay (y y0 ) v 2g(H y0 ) 2 y v02y 2g 2 0y H y0 2 0y v02y 2g [(22.4 m/s)(sin(35o)]2 H 114.7 m + 123. m 2 (2)(9.8 m/s ) v0 (x0, y0) H R x0 = 163.8 m y0 = 114.7 m 8 Tilted Rocket problem – part 3 (try on your own) A test rocket is launched by accelerating it along a 200.0-m incline at 1.25 m/s2 starting from rest at point A on the diagram. The incline rises at 35.0° above the horizontal, and at the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity (ignore air resistance).? What is the horizontal range of the rocket beyond point A? A) B) C) D) E) 280 m 339 m 357 m 402 m 621 m x0 = 163.8 m y0 = 114.7 m v0 v0 22.4 m/s (x0, y0) 9 To find total range R, determine time to fall to ground: 1 2 y(t) 0 y0 v0 y t gt 2 2 v0 y 2y0 t g g g v0y 2v0y 2y0 t t 0 g g 2 t 1.31 s (1.31 s)2 23.4 s2 6.32 s v0 (x0, y0) H R R x0 v0x t 163.8 m + [(22.4 m/s)(cos(35o)(6.32 s)] = 280. 10m Grinding wheel sparks problem What do you think the sparks flying off of a grinding wheel demonstrate most directly about the uniform circular motion? A. Velocity is tangential to the path B. Velocity is normal to the path C. Acceleration is tangential to the path D. Acceleration is normal to the path and directed towards the center of the circle E. Acceleration is normal to the path and directed away from the center of the circle Sparks fly off along the velocity vector of the particles and hence the velocity of the outer rim of the wheel Merry-Go-Round problem A boy and a girl ride on a merry-go-round. The boy rides on a “donkey” and the girl rides on a “horse”. The “donkey” is two times as far from the platform axis as the “horse”. What is the ratio of the centripetal acceleration experienced by the boy to the centripetal acceleration experienced by the girl? A. B. C. D. E. 1/4 1/2 1 2 4 2 v a R 2R v T 2 R 2 T 2 a = R T R 2 R doubles → v doubles → a doubles Shortcut: (see youtube video): a R Where ω is angular frequency: v 2 R T R doubles → a doubles 2 Hovering in mid-air problem How short would the day have to be in order for a person at the equator for gravity to no longer hold a person to the ground? The Earth’s radius is 6370 m. A) B) C) D) E) 1.4 minutes 14 minutes 1.4 hours 14 hours 140 hours a g R 2 2 g R T 2 R T 2 g T 2 (6.37 10 6 m)/(9.8 m/s2 ) 5100 s = 1.4 hr Neutron star problem A neutron star has a mass 1.4 times that of the Sun, but has a radius of only 10 km. As we will see later, that means its gravitational field at the surface is about 2 × 1011 Earth “g’s”. The fastest known rotation rate for a neutron star is 716 Hz. About how many of those 200 billion g’s are needed to keep an object on the equator from flying off? A) B) C) D) E) 2 thousand 2 million 2 billion 20 billion 200 billion Necessary centripetal acceleration: R 2 f R 2 2 [(2 )(716 Hz)]2 (104 m) = 2.0 1011 m/s2 ≈ 20 billion g’s