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Physics 140 – Fall 2014September 16

Reminder:

• Mastering Physics assignment 1 due tonight at midnight

Projectile & Circular Motion

wikimedia.org

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Questions on today’s youtube

video?

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?

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

Let’s find out!

Shooting a monkey

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2gt 2

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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) 11.2 m/sB) 22.4 m/sC) 33.6 m/sD) 44.8 m/sE) 56.0 m/s

v02 =2as

→ v0 = 2(1.25 m/s2 )(200. m) =22.4 m/sv0

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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).What is the maximum height above the ground that the rocket Reaches?

A) 67 mB) 94 mC) 123 mD) 156 mE) 199 m

x0 = (200.0 m) cos(35.0°) = 163.8 m

y0 = (200.0 m) sin(35.0°) = 114.7 m

(x0, y0)

v0 =22.4 m/s( )

v0

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(x0, y0)

v0

R

H

vy2 =0 =v0y

2 + 2ay(y−y0 ) =v0y

2 −2g(H −y0 )At the top:

→ H − y0 =v0y

2

2g→ H = y0 +

v0y

2

2g

→ H = 114.7 m +

[(22.4 m/s)(sin(35o)]2

(2)(9.8 m/s2 )= 123. m

x0 = 163.8 m

y0 = 114.7 m

What is the horizontal range of the rocket beyond point A?

A) 280 mB) 339 mC) 357 mD) 402 mE) 621 m

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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).?

x0 = 163.8 m

y0 = 114.7 m

v0 =22.4 m/s(x0, y0)

v0

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(x0, y0)

v0

R

H

To find total range R, determine time to fall to ground:

y(t) =0 =y0 + v0yt−

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gt2 → t 2 −2v0y

gt −

2y0

g= 0

t =v0y

g±

v0y

g

⎛

⎝⎜⎞

⎠⎟

2

+2y0g

t =1.31 s + (1.31 s)2 + 23.4 s2

=6.32 s

R =x0 + v0x

t=163.8 m + [(22.4 m/s)(cos(35o)(6.32 s)] = 280. m

What do you think the sparks flying off of a grinding wheel

demonstrate most directly about the uniform circular motion?

Grinding wheel sparks problem

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. 1/4

B. 1/2

C. 1

D. 2

E. 4

T

Rv

R

va

2

2

R doubles → v doubles → a doubles

→ a =

2π R

T⎛⎝⎜

⎞⎠⎟

2

R =

2π

T⎛⎝⎜

⎞⎠⎟

2

R

Shortcut: (see youtube video): a =ω 2R

ω =v

R=

2π

TWhere ω is angular frequency:

R doubles → a doubles

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) 1.4 minutes

B) 14 minutes

C) 1.4 hours

D) 14 hours

E) 140 hours

a =g=ω 2R

→ g =2π

T⎛⎝⎜

⎞⎠⎟

2

R

→ T = 2πR

g

→ T = 2π (6.37 ×106 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) 2 thousand

B) 2 million

C) 2 billion

D) 20 billion

E) 200 billion

Necessary centripetal acceleration:

ω 2R = 2π f( )2R

=[(2π )(716 Hz)]2 (104 m) = 2.0 ×1011 m/s2

≈ 20 billion g’s